Shoenfield's Mathematical Logic contains the following theorem and its corollary:
Closure Theorem. If ${\bf A}'$ is the closure of ${\bf A}$, then $\vdash {\bf A}'$ iff $\vdash {\bf A}$.
Corollary. If ${\bf A}'$ is the closure of ${\bf A}$, then ${\bf A}$ is valid in a structure ${\cal A}$ iff ${\bf A}'$ is valid in ${\cal A}$.
The proof of the corollary is presented as follows: Suppose that ${\bf A}$ is valid in ${\cal A}$. If $T$ has ${\bf A}$ as its only nonlogical axiom, then ${\cal A}$ is a model of $T$. By the closure theorem, $\vdash_T {\bf A}'$; so ${\bf A}'$ is valid in ${\cal A}$ by the validity theorem. The converse is proved similarly. $\square$
I can see intuitively that this is true, since the definition of validity in a structure is essentially asserting that the universally quantified version of ${\bf A}$ is valid (i.e. every instance of ${\bf A}$ is valid). Now, I don't completely understand why the case when ${\bf A}$ is the only nonlogical axiom of $T$ is sufficient for the proof. Note that at this point in the book, the Completeness Theorem is not yet proved and thus we can only infer that if a formula is a theorem, then it is valid (called the Validity Theorem in the book), but not the converse. Say that you had a theory $T$ with multiple nonlogical axioms or where ${\bf A}$ is not a theorem: how does the proof cover this case? Is it because the structure is applied to the language, rather than the theory itself?
Strictly speaking structure is applied to a language, model is applied to a theory as referenced here:
So if we have multiple nonlogical axioms $\bf A, B, C, \dots$ in $T$, the above proof holds since $\cal A$ can be always extended to be a model of $T$ satisfying all other axioms and sentences provided $T$ is consistent. The only possible issue is when $\bf A$ is neither a nonlogical axiom nor a theorem which is always correctly inferred from some axiom(s). But this seems impossible due to the specific definition of formula validity in this book mentioned on page 22: