Does Controlled-U gate entangle qubits?

For any controlled-$U$, if the input state is $|+rangle|phirangle$ where $|phirangle$ is not an eigenstate of $U$, then the output state is entangled. This immediately deals with trivial cases such as $U=I$ because in that case all states $|phirangle$ are eigenstates, and so it is not entangling. For any other $U$, there is an input state that is separable that is mapped to an entangled state, and hence controlled=$U$ is entangling for any $Uneq I$.

Proof: Let $|phirangle=sum_ia_i|lambda_irangle$, where $|lambda_irangle$ are the distinct eigenvectors of $U$. Your gate then evolves $$ |+rangle|phiranglemapstofrac{1}{sqrt{2}}|0rangle|phirangle+frac{1}{sqrt{2}}|1ranglesum_ia_ie^{ilambda_i}|lambda_irangle. $$ Since $sum_ia_ie^{ilambda_i}|lambda_irangle$ is not proportional to $|phirangle$ (under the assumption that the phases $e^{ilambda_i}$ are distinct and there are at least two non-zero $a_i$), the state is entangled because it is not a product state.