Capacitors on DC power rails / voltage divider - analyzing a circuit
Guitar gadgets are fun, aren't they? 


  
    C10/C11 seem to be part of low-pass filters to strip AC from the power lines
    why would we bother doing this if we assume the power supply is clean, or use a battery?
  


You answered your own question here but you don't realize it.   Key word "assume".   You are right, if the power source is an 'ideal voltage source' you would not need C10 or C11.  But it's not ideal.   There's no such thing.  It can bounce around a little due to a number of reasons (example, suppose you're jamming next to a radio transmitter...).  The caps make an effort to filter out noise.   Capacitors are cheap insurance. 


  
    how can C10 filter the +9v rail without a resistor? wouldn't the 9v just snap to the power source and bypass the capacitor?
  


The 'resistor' you don't see is in inside the power source because it is not ideal   ("output impedance").   


  
    My calculations put the C11/R13 low-pass filter very low, at 1/3 hz. I guess that's basically strict DC?
    assuming the op-amps take their power from the power rails, wouldn't this modify the voltage divider, making the DC offset change dynamically based on power consumption?
  


Great question and insight --- 

-- there's no power being consumed on the 4.5V rail.   It only goes to the input of the two opamps, and the inputs, for practical purposes, are "infinite resistance".  i.e. there's no load on the 4.5V line regardless whether you're taking a smoke break or rocking to an old Iron Maiden tune.
-- C10 keeps the 9V rail steady under any reasonable dynamic conditions and C11 keeps the 4.5 rail steady, even if C10 fails to fully filter the power to DC. 

The truth is, none of the rails will be truly DC ever.  The trick is to get them clean enough that the noise is not objectionable.  These circuits would not cut it on a NASA probe, but they work perfectly fine for guitar purposes.