I came across Los-Vaught's test:
Let $\mathcal{L}$ be countable and $\Sigma$ a $\mathcal{L}$-theory. $\Sigma$ is complete, if
$\Sigma$ only has infinite models.
There is a cardinal number $\kappa$ for which any two models of cardinality $\kappa$ are isomorphic.
In the proof we use 2. just to get elementary equivalence:
Let $A,B \in Mod (\Sigma)$, then $A \equiv A' \cong B' \equiv B$ hence $A \equiv B$.
Wouldn't it be enough to just ask for elementary equivalence in the second statement?
Yes, you are correct. We can replace "all models of size $\kappa$ are isomorphic" with "all models of size $\kappa$ are elementarily equivalent" and everything goes through fine.