My original question was the following:
Let $T$ denote an algebraic theory and suppose $X$ is a $T$-algebra. If for every identity $\eta$ in the language of $T$ we have that $(X \models \eta) \rightarrow (T \vdash \eta),$ is it necessarily the case that $X$ is a free algebra?
The answer is "no," because the first-order theory of $\mathbb{N}$ has non-standard models. In particular, let $T$ denote the equational theory of commutative unital semirings. Then $\mathbb{N}$ is the free $T$-algebra on $\emptyset$. Furthermore (binding the variable $\eta$ to the set of identities in the language of $T$), we have:
$$\forall \eta.(\mathbb{N} \models \eta) \rightarrow (T \vdash \eta).$$
However, there exist non-standard models of the first-order theory of $\mathbb{N}$. Let $M$ denote such a model. Then since $M$ satisfies the same identities as $\mathbb{N}$, we have that $M$ is a $T$-algebra. Furthermore:
$$\forall \eta.(M \models \eta) \rightarrow (T \vdash \eta).$$
However, $M$ is not isomorphic to $\mathbb{N}$, nor to any other free $T$-algebra.
My question now becomes, what happens if we assume $X$ is also finite?
Let $T$ denote an algebraic theory and suppose $X$ is a finite $T$-algebra. If for every identity $\eta$ in the language of $T$ we have that $(X \models \eta) \rightarrow (T \vdash \eta),$ is it necessarily the case that $X$ is a free algebra?
Let $T$ be the theory of Boolean algebras, and let $X$ be an $8$-element Boolean algebra (with $3$ atoms). Every identity satisfied by $X$ is satisfied by the $2$-element Boolean algebra (which is a subalgebra of $X$), and is therefore satisfied by all Boolean algebras, since every Boolean algebra is a subalgebra of a direct product of $2$-element algebras. However, $X$ is not free, since a free Boolean algebra on $n$ generators has $2^n$ atoms and $2^{2^n}$ elements.