Remember that if $G$ acts transitively on $X$, we must have $\vert X\vert \le\vert G\vert$. This sort of size bound runs counter to the compactness theorem:
Since $G$ is infinite, the compactness theorem - in its incarnation as the upwards Lowenheim-Skolem theorem - tells us that in order for the class of transitive $G$-sets to be elementary, there must be transitive $G$-sets of arbitrarily large cardinality. But the point above explains why that can't happen.
There are two differences between this answer and YCor's. First is the absence/presence of ultraproducts; this is really superficial (remember that ultraproducts provide a proof of the compactness theorem!). The second difference, which is meaningful, is that YCor's answer uses an even stronger fact about transitive $G$-sets than the cardinality bound above: that no transitive $G$-set has a proper extension. That is, if $\emptyset\not=X\subsetneq Y$, $G$ acts on $X$ and $Y$, the action of $G$ on $Y$ extends the action of $G$ on $X$, and $G$ acts transitively on $X$, then $G$ does not act transitively on $Y$.
Arguing via this stronger fact gives in some sense a more complete answer to the question, since it really shows just how far off from satisfying compactness the class in question is. By contrast, cardinality arguments like the above are rarely as satisfying, but do have the advantage of being broadly applicable and easy to use.
