Here is a little progress towards AC.
Theorem.
ICF implies the dual Cantor-Schröder-Bernstein
theorem, that is $X$ surjects onto $Y$ and $Y$ surjects onto $X$,
then they are bijective.
Proof. You explain this in the edit to the question. If
$X\twoheadrightarrow Y$, then $2^Y\leq 2^X$ by taking pre-images,
and so if also $Y\twoheadrightarrow X$, then $2^X\leq 2^Y$ and so
$X\sim Y$ by ICF. QED
Theorem. ICF implies that there are no infinite D-finite
sets.
Proof. (This is a solution to the exercise that you mention.) If $A$ is infinite and
Dedekind-finite, then let $B$ be the set of all finite
non-repeating finite sequences from $A$. This is also D-finite,
since a countably infinite subset of $B$ easily gives rise to a
countably infinite subset of $A$. But meanwhile, $B$ surjects onto
$B+1$, since we can map the empty sequence to the new point, and
apply the shift map to chop off the first element of any sequence.
So $B$ and $B+1$ surject onto each other, and so by the dual
Cantor-Schöder-Bernstein result, they are bijective,
contradicting the fact that $B$ is D-finite. QED
Here is the new part:
Theorem. ICF implies that $\kappa^+$ injects into
$2^\kappa$ for every ordinal $\kappa$.
Proof. We may assume $\kappa$ is infinite. Notice that
$2^{\kappa^2}$ surjects onto $\kappa^+$, since every
$\alpha<\kappa$ is coded by a relation on $\kappa$. Since
$\kappa^2\sim\kappa$, this means
$2^\kappa\twoheadrightarrow\kappa^+$ and consequently
$2^{\kappa^+}\leq 2^{2^\kappa}$, by taking pre-images. It follows that
$2^{2^\kappa}=2^{2^\kappa}\cdot 2^{\kappa^+}=2^{2^\kappa+\kappa^+}$ and
so by ICF we get $2^\kappa+\kappa^+=2^\kappa$, which implies
$\kappa^+\leq 2^\kappa$, as desired. QED
This conclusion already contradicts AD, for example, since AD
implies that there is no $\omega_1$ sequence of distinct reals,
which violates the conclusion when $\kappa=\omega$. In particular,
this shows that ICF implies $\neg$AD, and so in every AD model, there are sets of different cardinalities, whose power sets are equinumerous.
Best Answer
I've reduced the problem to "every infinite set is Dedekind infinite", which is a consequence of WPH if my memory serves me right (this was answered before on the site).
We'll show that well-ordered choice holds, which is known to imply Dependent Choice. In fact, we will argue the equivalent, $\forall A(\aleph(A) = \aleph^*(A))$.
Suppose that $A$ is an infinite set, then the following holds: $$\aleph(A) = \aleph(A\times\omega),$$ since $A$ is Dedekind infinite. And of course, $$|A\times\omega|=|A\times\omega|\cdot 2.$$
If $\kappa<\aleph^*(A\times\omega)$, then:
$$2^{A\times\omega}\leq 2^{A\times\omega}\cdot 2^\kappa = 2^{A\times\omega + \kappa}\leq 2^{A\times\omega}\cdot 2^{A\times\omega} = 2^{A\times\omega}.$$
By WPH, $|A\cdot\omega|=|A\cdot\omega+\kappa|$, so in particular, $\kappa<\aleph(A\cdot\omega)=\aleph(A)$.
Therefore, $\aleph(A) = \aleph^*(A\cdot\omega)$, and so we get $\aleph(A) =\aleph^*(A)$, so Dependent Choice holds.