David Pincus in "A note on the cardinal factorial" (Fundamenta Mathematicae vol.98(1), pages 21-24(1978)) proves that $A^A=2^A$ does not imply the axiom of choice, therefore it does not characterise the sets for which $A=A\times A$. The counterexample is the model from his paper "Cardinal representatives", Israel journal of mathematics vol.18,pages 321-344 (1974).
As Pincus writes on the last lines of [Pincus78]:
c. Our arguments [ in the proof of "4.The Main Theorem; If $x$ is infinite then $2^x=x!$" ] have made little use of the particular definition of $x!$. Indeed let $\mathcal{F}$ be any set valued operation which satisfies:
(1) The predicate $y\in\mathcal{F}$ is absolute from $M$ to $V$.
(2) $\mathsf{ZF}$ proves $|y|\leq x \Rightarrow |\mathcal{F}(y)|\leq |\mathcal{F}(y)|$ and $|2x|=|x|\Rightarrow 2^x\leq|\mathcal{F}(x)|$ for infinite $x$.
(3) $\mathsf{ZFC}$ proves $2^x=|\mathcal{F}(x)|$ for infinite $x$.
The statement "For every infinite $x$, $2^x=|\mathcal{F}(x)|.$", holds in $M$ (and therefore is not an equivalent to the axiom of choice). Examples of $\mathcal{F}$, apart from $x!$, are $x^x$ and $x^x-x!$.
Therefore, in this Pincus model, $\mathsf{ZF}$ + $\lnot\mathsf{AC}$ + "for all infinite x, $2^x=x!=x^x=x^x-x!=|\mathcal{F}(x)|$" holds for any set valued operator $\mathcal{F}$ as above.
I should say that, after failing to come up with an answer myself, I found out about this by searching into my good old friend the "Consequences of the axiom of choice" by Howard and Rubin, Form 200, and Note 64. I try to reference this book whenever I can :)