No such representation exists, unless the underlying spaces were trivial to begin with.
To see this let us first simplify notation by defining $\rho_{sa}:=U(\rho_s\otimes\rho_a)U^{-1}$ (which is obviously a state again). Now the question is: given $\{P_i\}_{i\in I}$ such that each $P_i$ is a positive semi-definite projection and $\sum_{i\in I}P_i={\bf1}_a$ when is $\{({\bf1}_s\otimes P_i)\rho_{sa}({\bf1}_s\otimes P_i)\}_{i\in I}$ a valid POVM? Our goal is to show that this is true (if and) only if both the system and the ancilla are one dimensional, hence trivial.
If $\{({\bf1}_s\otimes P_i)\rho_{sa}({\bf1}_s\otimes P_i)\}_{i\in I}$ were a valid POVM, then necessarily $\sum_{i\in I}({\bf1}_s\otimes P_i)\rho_{sa}({\bf1}_s\otimes P_i)={\bf1}_{sa}$. Using the projective property $P_i^2=P_i$, taking the trace yields
\begin{align*}
{\rm tr}({\bf1}_{sa})&={\rm tr}\Big(\sum_{i\in I}({\bf1}_s\otimes P_i)\rho_{sa}({\bf1}_s\otimes P_i) \Big)\\
&=\sum_{i\in I}{\rm tr}\big(({\bf1}_s\otimes P_i)\rho_{sa}({\bf1}_s\otimes P_i)\big)\\
&=\sum_{i\in I}{\rm tr}\big(({\bf1}_s\otimes P_i)({\bf1}_s\otimes P_i)\rho_{sa}\big)\\
&=\sum_{i\in I}{\rm tr}\big( ({\bf1}_s\otimes P_i)\rho_{sa} \big)\\
&={\rm tr}\Big( \Big({\bf1}_s\otimes \sum_{i\in I}P_i\Big)\rho_{sa} \Big)\\
&={\rm tr}(\rho_{sa})=1\,.
\end{align*}
In the last step we used that $\sum_{i\in I}P_i={\bf1}_a$ because $\{P_i\}_{i\in I}$ is a measurement, as well as that $\rho_{sa}$ is a state. But ${\rm tr}({\bf1}_{sa})$ is the product of the dimension of the system and the dimension of the ancilla, so this product being $1$ forces that both the system and the ancilla are of dimension $1$---hence they trivial, as we wanted to show.
