This is basically grounding one of the diff pair input.

1. Remember that the negative feedback will drive the amplifier to the voltage of the non-inverting input. That voltage needs to be set somehow.
2. You might think that it could be set internally, but the op-amp doesn't "know" where 0 V is relative to its supply rails. (Remember there is no GND pin on the op-amp, just positive and negative supply.) The supply could be single-ended, symmetrical or asymmetric.

Having the non-inverting input accessible gives the most flexibility.

The opamp differential input minimizes the input voltage offset, since any offset of one input's circuit is basically cancelled by the other input's identical circuit offset.

Also the differential input means the opamp can be used to make a variety of circuits that also use the (+) or both inputs (e.g. non-inverting amp, differential input amp, etc.).

It makes the opamp sort of the Swiss Army Knife of analog circuits

The inertia of verbal clichés

I understand your insistent desire to get an answer to this fundamental question, because I have thought a lot about this too. Just take a look at the list of links at the end to get an idea of how great the interest in the differential arrangement is and what effort it has cost me to show what the idea behind it is. Human thinking tends not to delve deep but to use formal, worn-out explanations, because this is easier and yields fast results. It is a fortunate circumstance that there are people like you who are not satisfied with such superficial explanations and persistently seek the idea, which motivates us to find a worthy answer to such a profound question.

Short answer

Is there still any practical benefit of the differential pair, and what noise/offsets is the CMRR rejecting?

The answer depends on what kind of op-amp you have in mind in this inverting amplifier circuit.

• If this is a conceptual circuit with an "ideal" op-amp, the answer is, "No, there is no benefit from an amplifier with a differential input. Its use is meaningless and only complicates the circuit."

• However, if this is a practical circuit with a real op-amp, which is subjected to certain harmful effects, the answer is, "Yes, there is a benefit from a differential input stage, because it will compensate for them."

But how does this "magical" suppression of harmful effects and amplification of useful effects happen?

Basic idea

The idea is ingeniously simple. Instead of using just one single-ended amplifier, we take another (single-ended) amplifier, place it close to the first one, and connect it in the opposite direction.

​In this way, all "common" effects (which act simultaneously on both amplifiers) destroy each other, and the output voltage does not change. Examples include the supply voltage, temperature, the transistors' beta, and others.

However, the input voltages play a more specific role—since we have applied them to two separate inputs, they can be both the same (common mode) and different (differential mode). This finds application in transmitting signals over long distances and in the non-inverting op-amp amplifier.

In the inverting amplifier, however, the input voltages are used only in differential mode. For this purpose, one input is fixed to ground or a fixed potential, and only the other input is used. Thus, the op-amp is sensitive to (responds to, amplifies) only one input voltage; it has, as you call it, a "single-ended input".

However, all other common effects (the supply voltage, the temperature, the transistors' beta, etc.) remain compensated and do not affect the output voltage. This is why there is a point in using an op-amp with a differential input even in the case of an inverting amplifier.

The essential thing here is to understand that the common mode of the input voltages has nothing to do here, and the classic example of transmitting a signal over a distance is not relevant. Only the other quantities (including non-electrical ones) remain, which always act in common mode.

Related answers

Inverting configuration

Conception of an inverting op-amp with a specific gain

Differential configuration

Is the voltage on the negative of a differential pair actually below the ground voltage used to generate the signals?

Transistor implementation

Designing a differential amplifier

What is the best way to analyze this differential amplifier?

Questions about a differential emitter follower

What is the idea behind the transistor differential amplifier?

How does the current and voltage stabilize in a differential amplifier?

BJT Differential Amplifier problem

How does an op-amp know where ground is?

Op-amp implementation

Ground in differential opamp situation

Differential Amplifier with Unity Gain Source vs Instrumentation Amplifier

When is it an instrumentation amplifier (In-Amp) and not an operational amplifier (Op-Amp)?

What is the idea behind the op-amp instrumentation amplifier?

Is a common ground need in a differential amplifier?

Every configuration of an op amp uses the input as a differential pair. The only equation for every op-amp configuration is \$V_{\text{out}}=A(V_+-V_-)\$, where \$A\$ is a very large number.

In negative feedback situations, like your simple inverting amplifier, we arrive at the output equations by asserting the limit as \$A\rightarrow \infty \$, which is usually fair so long as we're not doing anything that would violate that assertion, such as asking for too much gain.

There are configurations of op amp circuits that provide outputs proportional to the differences between two inputs. Those all apply the same equation to \$V_+\$ and \$V_-\$ that every other op amp circuit does.

Yes, the latter differential amplifiers have a better tendency to remove common mode noise than non-differential amplifiers, for a variety of reasons (See Twisted-pair wire's effect on Common mode noise and Andy's answer for more thorough description), but the way you pose the question is really conflating two issues and causing confusion -- there isn't any real reason to conflate the differential inputs of an op-amp and the differential inputs of an op-amp circuit.

To make an Op-Amp (with this, I mean, an amplifier with a large open loop gain with a feedback network), a single ended stage can be, many times, sufficient. An exception to this can be if you're trying to amplify a DC signal, in which case you can't tolerate the 1st stage giving you a DC offset, like a NMOS whose (-) terminal will be its source.

Balancing techniques, like using a diff. pair can have nice benefits like increased common-mode rejection, for example, but they are not a must.

I've designed several custom IC amplifiers with single-ended inputs due to tough noise requirements and headroom issues and validated them in the lab with excellent noise ans distortion specs.

If I understand the question correctly, I actually did that in a small off-the-shelf analog mixing console, to move the USB input from the almost-useless "tape" input to a full stereo channel strip.

That channel strip has two 1/4" TRS jacks that are normalled to ground, which means that the jack inputs are literally and intentionally shorted to ground when there's no plug in them, so I can't just tap into those. (yes, I could use a dummy plug, just to open the switch, but you KNOW that someone is going to unplug that and lose it...)

So I found the single-opamp-per-channel diff-amp that sits behind each of those jacks, and stuck a 13k 0805 resistor from my grab bag, vertically on the inverting pins of each opamp channel (intentionally tombstoned, if you will), and then stuck a wire on that to go to the USB board. The on-board resistors for that diff-amp are 10k, so this guarantees that the USB input can never clip, assuming a similar signal level as the jack expects.

Works perfectly! Though the USB input is now inverted, which violates convention, but where it's going, no-one's going to notice.

^{simulate this circuit – Schematic created using CircuitLab}

Thus, the opamp's (+) input stays at ground potential, with nothing to disturb it. It has 5k of impedance to reject noise with (10k||10k), compared to a dead short to ground as is normally done for an inverting-only configuration, but the normal factory performance is also "limited" by that same thing, so it's no different from that.

Now, the opamp will do whatever it needs to, to hold its (-) input at the same potential. From here, we can consider currents, assuming that the opamp is successful in that goal. R1 always has 0V on both sides, so no current there, and so it effectively disappears. Now we only have R5 and R2 in the standard inverting configuration as usual. Standard analysis from there.

Is a differential pair stage of an opamp still useful with single ended input

It certainly is because there are many applications where a single-ended input hasn't "collected" noise and, can be perfectly amplified by a simple inverting or non-inverting op-amp circuit.

Is there still any practical benefit of the differential pair, and what noise/offsets is the CMRR rejecting?

Well, it is a differential amplifier and, it can deal with noise/offsets providing they affect both the single-ended signal and 0 volts together. Of course, with short connections on a PCB that utilizes a ground/0 volt analogue plane then this is hardly an issue. I mean if we get what some folk will refer to as "ground-bounce" it will equally affect signals and 0 volts simultaneously so, there is no problem or deterioration of signal-to-noise ratio.

The main area where you do need to use a fully differential input op-amp is in the rejection of noise that affects the signal and its return on the path they take to a receiver. And the reason we use a differentially configured op-amp circuit is to provide balanced impedances to the signal and its return path. But, both the receiver and the transmitting circuit need to have a balanced impedance or this happens: -

Scenario B uses an extra resistor at the transmitting end and this provides the solution. Image from my answer here.

See also Differential vs non-inverting op amp advantage.

"Grounding" one input of an op-amp is the best you can do to maintain said op-amp's common mode rejection performance. The common-mode - in this case the (+) input - is held constant. Since the common mode voltage doesn't change, the CMRR doesn't matter - there's no common mode to reject.

So, if anything, the inverting configuration - everything else being the same - is better if the signal source can accommodate the load of the feedback network. The inverting configuration lets you use an op-amp with poorer CMRR specification - be it for cost reduction, or for the flexibility in choosing an op-amp that has better some other specification of interest but trades off poorer CMRR for it.

Modern op-amps perform so well that it is extremely hard to design an input feedback network that is symmetric enough to let one operate the op-amp as a differential-to-single-ended converter. Said network needs to have matched DC and AC impedances, and equally well matched must be the parasitic impedances of the traces on the PC board and the leadframe/bonding of the op-amp die.