Let $T$ be a first-order theory over a language $L$, and let $\mathcal{M}$ be a subclass of the class of models of $T$. As I understand it, if there is no theory $\hat{T}$ over $L$ whose class of models is exactly $\mathcal{M}$, frequently, the "morally correct" reason is that $\mathcal{M}$ is not closed under ultraproducts. However, it is sometimes possible to obtain a simpler proof of non-axiomatisability by compactness or completeness considerations.
Example. Consider the first order theory of fields, and let $\mathcal{M}$ be the class of fields of positive characteristic. Then $\mathcal{M}$ is not axiomatisable in the language of fields since, for example, if we take the ultraproduct of all the finite fields $\mathbb{F}_p$, $p$ prime, we would obtain a field of characteristic 0.
We could also prove this directly: if T' is the naïve axiomatisation of fields of characteristic 0 (i.e. the one with one axiom of the form $\underbrace{1 + \cdots + 1}_{n\text{ times}} \ne 0$ for every positive $n \in \mathbb{N}$), and $\hat{T}$ is any axiomatisation of $\mathcal{M}$, then T' \cup \hat{T} is inconsistent, so there is some finite subset which is inconsistent, so there is some finite set $X$ for which $\hat{T}$ proves that there is an $n \in X$ such that $\underbrace{1 + \cdots + 1}_{n\text{ times}} = 0$; but there are fields of positive characteristic other than those $n \in X$ — a contradiction.
Question. Is it in fact always possible to translate a proof of non-axiomatisability using ultraproducts to one using compactness/completeness?