Does two quantum states interacting via some quantum interaction, always gets entangled? We know that two quantum states, can not get entangled via local classical interactions/communication (LOCC). However, do two quantum states locally interacting via quantum interactions always get entangled?
Suppose, two Quantum systems, initially in the unentangled state, interact locally with a mediator, and no entanglement is generated between the two because of interaction, can we unambiguously determine that the mediator was classical in nature?
 A: Answer
However, do two quantum states locally interacting via quantum interactions always get entangled?
No. I can give you an example using the $CNOT$ gate for qubits which is defined by the following relation:
$$ CNOT |0\rangle |0\rangle = |0 \rangle |0\rangle $$
$$ CNOT |0\rangle |1\rangle = |0 \rangle |1\rangle $$
$$ CNOT |1\rangle |0\rangle = |1 \rangle |1\rangle $$
$$ CNOT |1\rangle |1\rangle = |1 \rangle |0\rangle $$
i.e. the second qubit is flipped if the first one is in state $|1\rangle$. Clearly none of the above states is entangled after operating with CNOT.This is because the state can be written in the form $|\cdot\rangle_{q_1}|\cdot\rangle_{q_2}$ where $q_1$ and $q_2$ are qubits 1 and 2.
Now, look at the not-entangled state $$\frac{1}{\sqrt 2}(|0\rangle + |1\rangle)\otimes |1\rangle = \frac{1}{\sqrt 2} (|0\rangle|1\rangle + |1\rangle|1\rangle)$$
after acting with $CNOT$ we obtain:
$$CNOT \frac{1}{\sqrt 2} (|0\rangle|1\rangle + |1\rangle|1\rangle) = \frac{1}{\sqrt 2} (|0\rangle|1\rangle + |1\rangle |0\rangle)$$
using the defining relations of the CNOT gate. The state $\frac{1}{\sqrt 2} (|0\rangle|1\rangle + |1\rangle |0\rangle)$ is called Bell state and it is a maximally entangled state. Hence, you can see that the final level of entanglement also depends on the initial state that experiences the interaction.
Extension: Entangling Power
After doing some more reading, I came across the concept of entangling power. For a gate $U$, the entangling power $K$ is defined as [1]:
$$K_{E}(U)=\max _{|\phi\rangle,|\psi\rangle} E(U(|\phi\rangle|\psi\rangle))$$
This quantity is the maximum attainable entanglement $E$ maximised over all states $|\phi\rangle|\psi\rangle$.
[1] https://iopscience.iop.org/article/10.1088/1751-8121/aad7cb/pdf
