Do linear operators $A$, $B$ satisfying $A = B+BAB$ commute? I have two linear continuous operators $A$, $B$ on Banach space $X$ (for example, square matrices), satisfying the equation
$$
   A = B + BAB,
$$
and such that the continuous inverses $(\mathrm{Id} \pm B)^{-1}$ exist. Is it true that in this case $AB = BA$? In the case when $\|B\|<1$
 it follows from the following series expansions for $A$:
$$
  A = B + B^3 + B^5 + \cdots.
$$
P.S. The similar result holds, for example, if operators $A$, $B$ satisfy the equation
$$
   A = B + AB,
$$
and there exists continuous inverse $(\mathrm{Id}-B)^{-1}$. 
In this case we can show that operators $P = AB$ and $Q = BA$ satisfy the equations
$$
   P = B +PB, \\
   Q = B +QB,
$$
and hence $(P - Q)(\mathrm{Id}-B) = 0$ and $P=Q$.

Counterexample: $B = \left(\begin{array}{cc} 2 & 0 \\ 0 & 1/2 \end{array}\right)$ and $A = \left(\begin{array}{cc} -2/3 & 1 \\ 1 & 2/3 \end{array}\right)$.
