Hi again,
This looks a lot like some of the formula's you get for propagators in QFT for some background field calculations. (Combined with your derivative of a determinant question, the evidence is even stronger). Of course, in QFT the inner products are infinite dimensional, not finite.
I think that you might find that this will be impossible to get a closed form for this (though I'd like to be proven wrong). If you assume various commutativity properties then you might be able to do better.
Assuming the G's are invertible, I'd do something like
0 = G (A + G' M-1 G)-1 G' + F + M
0 = G G-1(G'-1 A G-1 + M-1)-1 G'-1 G' + F + M
0 = M (1 + G'-1 A G-1 M)-1 + F + M
M = -F (1 + (1 + G'-1 A G-1 M)-1)-1
This gives a continued fraction expansion for M that can be taken to any order that you need. You could also write down the exact answer using a noncommutatitve continued fraction notation.
In QFT if you expand the continued fraction, you get an expression that looks something like mass insertions:
perturbative propagator = -- = -F
mass 2-point function = x = G'-1 A G-1
exact propagator = --o-- = M = -- + --x-- + --x--x-- + --x--x--x-- + ...