In this question the answer by Alex Kruckman gives a standard argument using the compactness theorem for why a torsion group with elements of arbitrary large order is elementarily equivalent to a nontorsion group. Paraphrasing slightly:
Let $G$ be a torsion group containing elements of arbitrarily large order. Let $c$ be a new constant symbol, and consider the theory $T' = \text{Th}(G)\cup \{c^n \neq e\mid n=1,2,...\}$ (where $c^n$ is an abbreviation for the product of $n$ copies of $c$). By compactness, this theory has a model, which is a non-torsion group elementarily equivalent to $G$ (given a finite subset of $T'$, take $G$ and interpret $c$ as an element of large enough order).
I am having trouble seeing why the resulting group, call it $H$, is elementarily equivalent to $G$ (ignoring the new constant $c$ so that $G$ and $H$ are two structures in the same first order language $\mathcal L$). Clearly every sentence true in $G$ is true in $H$ since $H$ is a model of $\text{Th}(G)$. For the converse, given a sentence $\phi$ in $\mathcal L$ holding in $H$, the constant $c$ is not in the language $\mathcal L$, so does not appear in $\phi$. But why does $\phi$ also hold in $G$?