The proof that I know of the class $C$ of finitely chromatic (possibly infinite) graphs not being closed under elementary equivalence involves ultraproducts as in the following. First, one can show that if $C$ is closed under ultrapowers then it is also closed under ultraproducts. Indeed, if $G_i \in C$ for $i \in I$, then the ultraproduct of $(G_i)_{i \in I}$ modulo $D$ can be embedded as a connected component of the ultrapower modulo $D$ of the disjoint union of $(G_i)_i$, which is finitely chromatic by assumption. Then one proceeds by showing that $C$ is not an elementary class. This is done by, for example, taking the ultraproducts of $(G_i)_{i < \omega}$, which is a sequence of finite graphs of arbitrarily large girths and arbitrarily large chromatic numbers $< \omega$.
Can one show the same thing without using ultraproducts? That is, can I replace the argument with one that uses Loewenheim-Skolem, realizing/omitting types, etc.?
Addendum: bof has provided the proof of $C$ actually closed under $\equiv$, which I understand. I'm confused; what's wrong in the argument above?
As mentioned in the comments, this is not true. The error in your argument is that $C$ is not closed under infinite disjoint unions: if you have a family of graphs $G_i$ such that $\chi(G_i)$ is finite for each $i$ but not bounded, then the disjoint union will have infinite chromatic number. So in fact $C$ may be closed under ultrapowers but not under ultraproducts, and this is in fact what happens (and correspondingly, it is closed under elementarily equivalence but is not an elementary class).