Suppose we have a Borel function $f:\mathbb{R}^n\to\mathbb{R}^n$ that is bijective such that its inverse is also Borel. Assume that every finite measure $\mu$ on $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n))$ with density $h_\mu$ with respect to Lebesgue measure satisfies that its push-forward $f_\sharp\mu:=\mu\circ f^{-1}$ also admits a density $h_{f_\sharp\mu}$ with respect to Lebesgue measure.
Question: can we say something about the differentiability of $f$? My intuition is that $f$ should be a.e. differentiable -- sort of a converse of the change of variable formula.
We unfortunately cannot say that $f$ will be continuous even at a single point, let alone differentiable.
First note that your condition is equivalent to the condition that $f$ and $f^{-1}$ satisfy Luzin's condition $N$ for Borel sets, i.e., they send Borel sets of Lebesgue measure $0$ to Borel sets of Lebesgue measure $0$. Indeed, certainly such maps satisfying condition $N$ will push absolutely continuous measures forward to absolutely continuous measures (otherwise they would map some set of positive Lebesgue measure onto a Lebesgue nullset).
Conversely, say you are given a map $f$ that preserves the condition of absolute continuity with respect to Lebesgue measure under pushforwards. Let $S\subseteq \mathbb R$ have finite and positive Lebesgue measure. Then $f_{\#}m\llcorner_{S}$ will have positive measure and be concentrated on $f(S)$, so absolute continuity of $f_{\#}m\llcorner_S$ implies $f(S)$ has positive Lebesgue measure as well.
However, we can construct an everywhere discontinuous bijection $f$ with condition $N$ for $f$ and $f^{-1}$ as follows.
We will instead of $\mathbb R$ use $(0,1)$. Let $S\subseteq (0,1)$ be a Borel set such that $m(S\cap I)>0$ and $m(I\backslash S)>0$ on every interval $I\subseteq (0,1)$. (See here for an explicit construction of such a set). Note that we may construct $S$ in such a way that $ x\in S$ if and only if $1-x\in S$ (the linked construction satisfies this if done symmetrically).
Then define $$f(x)= \begin{cases} x & x\in S\\ 1-x & x\in (0,1)\backslash S. \end{cases} $$
Then $f=f^{-1}$ is certainly a Borel bijection. Moreover, $f$ satisfies condition $N$, which can be seen by restricting $f$ to $S$ and $(0,1)\backslash S$ separately.
However, $f$ is not continuous at any point except $x=\frac{1}{2}$, which can be remedied by manually switching the value of $f$ at $x=\frac{1}{2}$ with any the value of $f$ at any other point. (Said switching will never accidentally make $f$ continuous elsewhere, since none of the discontinuities are removable.)
Remark
Of course, we could have simply taken $S=\mathbb Q\cap (0,1)$ and accomplished the same task, i.e., we only technically needed density of $S$ and $(0,1)\backslash S$, but the preceding example shows that we can construct our bijection so that it cannot be altered on a set of measure $0$ to be continuous.