Let ${\cal L}$, ${\cal L'}$ be first order languages.
Let $T'$ be an ${\cal L}'$ theory, and let ${\bf C}$ be the class of all ${\cal L}$ structures which can be expanded to a model of $T'$.
Show that if ${\bf C}$ is closed under taking substructures then ${\bf C}$ is universally axiomatizable.
Consider the theory $T$ consisting of all universal ${\cal L}$ formulas $\phi$ with $T' \models \phi$.
Suppose ${\frak A} \models T$. We claim that $T' \cup {\rm Diag}({\frak A})$ is consistent.
If not, there is a formula $\phi(a) \in {\rm Diag}({\frak A})$ such that $T' \vdash \neg \phi(a)$. But then since $\phi$ is quantifier free, $T' \vdash \forall x \neg \phi(x)$, so $\forall x \neg \phi(x) \in T$, a contradiction since ${\frak A} \models \exists x \phi(x)$.
Let ${\frak C}$ be a model of $T' \cup {\rm Diag}({\frak A})$. Then ${\frak A}$ embeds into ${\frak C}$. So ${\frak A} \in {\bf C}$ since ${\bf C}$ is closed under substructures.
Conversely, if ${\frak A} \not \models T$, then clearly we can't have ${\frak A}' \models T'$ for an expansion of ${\frak A}$.
So $T$ is a set of universal axioms for ${\bf C}$.