Taking advantage of the model theory, prove that $M \cong N \implies M \equiv N$
HINT: Let $h:M\to N$ be an isomorphism. Show by induction on the complexity of $\varphi(x_1,\dots,x_n)$ that $M\vDash\varphi(a_1,\dots,a_n)$ iff $N\vDash\varphi\big(h(a_1),\dots,h(a_n)\big)$. (By now you should have seen several such proofs.) From this it follows immediately that $M\equiv N$.