Is there an entire function with domains for which $f(A)=B$ and $f(B)=A$?

The conclusion does not hold, not even for polynomials. If $z_0$ is an attracting fixed point of $f \circ f$ (but not a fixed point of $f$) and $A$ the component of the Fatou set containing $z_0$, then $B = f(A)$ is disjoint from $A$ with $f(B) = A$.

A concrete example is $f(z) = z^2 - 1$ with $f(0) = -1$, $f(-1) = 0$, and $A, B$ the components of the Fatou set containing $0$ and $-1$, respectively.

Here is an image of the Julia set of $z^2-1$ (Attribution: Prokofiev / Public domain):

The Fatou component in the center contains $z=0$ and the next one on the left contains $z=-1$.