 Solution

No: $U$ is not an entangling gate. Its factorization gives, for every product input,which remains a product state. More generally, a convex mixture of product density matrices remains separable under these local unitary operators. Thus being distinguishable from $I\otimes V$ does not require creating entanglement. The first qubit undergoes a genuine local phase gate, while the target undergoes the same $V$ in both branches.