This is only a partial answer because I'm having trouble reconstructing something I think I figured out seven years ago...
It would seem the Dual Cantor-Bernstein implies Countable Choice. In a post in sci.math in March 2003 discussing the dual of Cantor-Bernstein, Herman Rubin essentially points out that if the dual of Cantor-Bernstein holds, then every infinite set has a denumerable subset; this is equivalent, I believe, to Countable Choice.
Let $U$ be an infinite set. Let $A$ be the set of all $n$-tuples of elements of $U$ with $n\gt 0$ and even, and let $B$ be the set of all $n$-tuples of $U$ with $n$ odd. There are surjections from $A$ onto $B$ (delete the first element of the tuple) and from $B$ onto $A$ (for the $1$-tuples, map to a fixed element of $A$; for the rest, delete the first element of the tuple). If we assume the dual of Cantor-Bernstein holds, then there exists a one-to-one function from $f\colon B\to A$ (in fact, a bijection). Rubin writes that "a 1-1 mapping from $B$ to $A$ quickly gives a countable subset of $U$", but right now I'm not quite seeing it...
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
Yes. This is a combination of facts.
$\sf PP$ implies that if a set $X$ can be mapped onto an ordinal $\alpha$, then $\alpha$ injects into $X$. In other words, it implies that $\aleph^*(X)=\aleph(X)$ for any set $X$.
$\sf AC_{WO}$, that is the axiom of choice from families of sets indexed by an ordinal, is equivalent to the statement "For every $X$, $\aleph^*(X)=\aleph(X)$".
$\sf AC_{WO}$ implies $\sf DC$. This is due to the fact that if $T$ is a tree of height $\omega$ without cofinal branches, we can consider the various well-orderable subtrees of $T$, define a rank function on those and use that to define a rank function on $T$. Then, using $\sf AC_{WO}$ we can show that it is impossible that $T$ itself is not truly well-founded, which would mean that it has a maximal element.