In theory, you might be right, but what happens if one of them is 22u—20% and the other is 22uF+20%?
What happens if one capacitor has more leakage current than the other?
What effect does the IC input bias current have?
And how are capacitive dividers supposed to work on DC?
It’s a pity that simulators are not idiot-proof.

Mismatched resistors are just as bad as mismatched capacitors. Tight tolerance capacitors do exist. While the overall circuit sees the two 22uf capacitors in series, as an 11uF power supply bypass capacitor, the non inverting input of A2, sees a 44uF capacitor, with respect to ground.

 

Capacitive dividers work similar to resistive dividers, however, I wouldn't expect much output current from such low value capacitors.

That doesn' work very well as the charge begins to slowly increase past the 4.5 volt reference.
Verified on the breadboarded with several different values of capacitors. Two 100uf seemed the best but still creeping slowly.
Just use two resistors with a capacitor at the center, much simpler and stable.

 

Don't forget to use a 10uF capacitor across the V+ and V- pins.

 

THE TWO RESISTORS CAN BE ONE MEGOHM and that can work with the output already capacitor coupled and the unbalance may eventually cause a problem but possibly the clipping will not be noticed, depending on the kind of music.

 

That doesn' work very well as the charge begins to slowly increase past the 4.5 volt reference.
Verified on the breadboarded with several different values of capacitors. Two 100uf seemed the best but still creeping slowly.
Just use two resistors with a capacitor at the center, much simpler and stable.

I this what you have in mind ?

 

Here's another configuration using the TL082.
The first unity gain stage provides a virtual common at the output eliminating the need for the capacitors.

 

That's interesting.

 

In actual practice I needed to change R1 to 39K for a symmetrical output when clipping. YMMV.

 

That makes sense, considering that the output can swing up to 8.1 volts, and down to 1.8 volts. Ideal bias voltage = ( 8.1 + 1.8 ) / 2 = 4.95 volts.