Non existence for "forgetful" measurements
Suppose after the entangling operation the state is some $\rho_{XY}$. Given a POVM $\{M_a\}_a$ we can describe the final state of system $Y$ and its correlations with the measurement outcome as the classical-quantum state
$$
\rho_{AY} = \sum_a |a \rangle \langle a| \otimes \mathrm{Tr}_X[(M_a \otimes I_Y) \rho_{XY}].
$$
If we "do not read the measurement outcome" then we can just look at the marginal state on $Y$
$$
\mathrm{Tr}_A[\rho_{AY}] = \sum_a \mathrm{Tr}_X[(M_a \otimes I_Y) \rho_{XY}] = \mathrm{Tr}_X[\rho_{XY}].
$$
But now note that if $\rho_{XY}$ is entangled, it is impossible for $\mathrm{Tr}_X[\rho_{XY}]$ to be a pure state. Thus the update on the system $Y$ cannot be unitary.
Existence of a unitary transformation simulating $Y$ update
Looking at a slightly weaker question, could there exist a unitary transformation that mapped the initial state to the final state then if you condition on the measurement outcome, i.e., you look at the state
$$
\rho_Y(a) = \mathrm{Tr}_X[(M_a \otimes I_Y) \rho_{XY}]
$$
it is indeed possible that $Y$ ends up in a pure state. Just take $\rho_{XY}$ to be any pure entangled state and take $M_a$ to be any rank-one measurement then $\rho_{Y}(a)$ will be pure and hence there will exist a unitary transformation (depending on the measurement outcome) mapping between the initial state and the final state.
Note this does not mean that the transformation you describe necessarily implements a unitary on $Y$.
Example of when $Y$ update is a unitary conditioned on the measurement outcome
To answer the original question in the positive consider the unitary mapping
$$
U = |0 \rangle \langle + |\otimes I + |1 \rangle \langle - | \otimes X
$$
This state has the action $U|00\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}$. If you then measure in the computational basis you find
$$
\rho_Y(0) = |0 \rangle \langle 0| \qquad \rho_Y(1) = |1 \rangle \langle 1|
$$
which corresponds exactly to the action of the mapping on $Y$ which is, apply an identity $I$ when the outcome is $0$ and apply a Pauli $X$ when the outcome is $1$. You can verify this directly by computing
$$
\mathrm{Tr}_X[(|0 \rangle \langle 0| \otimes I) U (\rho_X \otimes \rho_Y) U^{\dagger} (|0\rangle \langle 0| \otimes I)] = \langle + | \rho_X |+\rangle \rho_Y
$$
and
$$
\mathrm{Tr}_X[(|1 \rangle \langle 1| \otimes I) U (\rho_X \otimes \rho_Y) U^{\dagger} (|1\rangle \langle 1| \otimes I)] = \langle - | \rho_X |-\rangle X\rho_Y X
$$
