For a density matrix $\rho_{AB}$ and some operators $A, B$, is there a way to express
$$\text{Tr}_A((A\otimes B)\rho_{AB})$$
using the reduced states $\rho_A$ and $\rho_B$ and operators $A$ and $B$? Note that when $A$ is the identity operator, this is indeed possible and one has
$$\text{Tr}_A((I_A\otimes B)\rho_{AB})= B\rho_{B}$$
as shown here.
In general, the knowledge of the marginals $\rho_A$ and $\rho_B$ and the operators $A$ and $B$ is insufficient to compute $\mathrm{tr}_A((A\otimes B)\rho_{AB})$. Indeed, we can find two different density matrices $\rho_{AB}$ and $\sigma_{AB}$ with the same marginals for which
$$\mathrm{tr}_A((A\otimes B)\rho_{AB}) \ne \mathrm{tr}_A((A\otimes B)\sigma_{AB}).$$
One such counterexample uses the elements of the Bell basis $|\beta_{ij}\rangle = X_A^iZ_A^j|\beta_{00}\rangle$ with $|\beta_{00}\rangle=(|00\rangle+|11\rangle)/\sqrt{2}$ and $i,j=0, 1$. In this case
$$ \begin{align} \mathrm{tr}_A((A_A\otimes B_B)|\beta_{ij}\rangle\langle\beta_{ij}|) &= \mathrm{tr}_A\left[(A_A\otimes B_B)X_A^iZ_A^j|\beta_{00}\rangle\langle\beta_{00}|Z_A^jX_A^i\right] \\ &= \mathrm{tr}_A\left[(A_A\otimes B_B)Z_B^jX_B^i|\beta_{00}\rangle\langle\beta_{00}|X_B^iZ_B^j\right] \\ &= \mathrm{tr}_A\left[(I_A\otimes B_B)Z_B^jX_B^iA_B^T|\beta_{00}\rangle\langle\beta_{00}|X_B^iZ_B^j\right] \\ &= BZ^jX^iA^T\,\mathrm{tr}_A\left[|\beta_{00}\rangle\langle\beta_{00}\right]\,X^iZ^j \\ &= \frac12BZ^jX^iA^TX^iZ^j \end{align}\tag1 $$
where we exploited a useful and easy to check property that $(A\otimes I)|\beta_{00}\rangle = (I\otimes A^T)|\beta_{00}\rangle$. We see that if $A$ and $B$ are non-singular, then for the expression $(1)$ to yield the same value for all $i,j$ we need $A$ to commute with $X$ and $Z$. However, any operator $A$ on $\mathbb{C}^2$ can be written as $A=aI+bX+cZ+dXZ$ and it is easy to see that if $A$ commutes with $X$ and $Z$ then $b=c=d=0$. Consequently, $A=aI$.
We conclude that if $A\ne aI$ then $\mathrm{tr}_A((A\otimes B)\rho_{AB})$ is not uniquely defined on the set of density matrices $\rho_{AB}$ with given marginals $\rho_A$ and $\rho_B$. Therefore, no general formula for $\mathrm{tr}_A((A\otimes B)\rho_{AB})$ in terms of $A$, $B$ and the marginals exists.
No - take $\rho$ one of the four Bell states, which all have the same marginals. Then the trace you give will evaluate to $\mathrm{tr}[APBP]$, with $P$ one of the four Paulis (including $I$), which is not only a function of $A$ and $B$ (as it depends on the Pauli, which cannot be inferred from the reduced states).
