Electronic – Why change the quiescent current in an op-amp? (LM4250)

It's hard to tell what you are asking because some things you say don't make sense. In the schematic you show, both outputs are clearly referenced to ground. Pin 5 produces the inverted signal and pin 8 the non-inverted, but both are referenced to ground. The DC offset of both will be close to half the supply voltage.

However, if the point of this block is to be a preamp, then the choice of this part makes no sense. You didn't supply a link to the datasheet so I don't know what a LM4871 is exactly, but from the schematic you show it appears to be a small power amp. You can see that the gain of each output is 1 referenced back to the input. The main job of this circuit seems to be to provide power gain (lower impedance output) as apposed to voltage gain, and to produce a differential signal to drive a small speaker with directly. This does actually give you a voltage gain of 2 accross the speaker relative to the input, but the main point is to be able to drive the speaker with ± the supply voltage. That is useful to put more power into the speaker from a low supply voltage.

If you want a preamp, make a preamp, and use parts intended for some voltage gain with good characteristics of noise, linearity, and distortion. This is not what the LM4871 is intended for, given the schematic you provided.

No, you should not isolate the grounds of the two op-amps and then directly connect the output of one to the input of the other.

Imagine in your circuit if DGND were 50 V above (or below) AGND. Then the input to U1 would be well outside the common mode operating range. Realistically, the connection of the circuit would prevent the two grounds by drifting so much. But it would only stop them by allowing excess current to flow, resulting in unexpected circuit behavior.