Let $K$ be the class which contains all structures in the union of all classes of undirected graphs which are not finitely axiomatizable
I noticed that for example there are arbitrrily infinite many of such classes, for example one can look at the classes of infinite graphs which are distinguished by the existence of a node that has $m$ edges where $m\in M$ where $M$ is an arbitrary set. So I cannot use Skolem to say $K$ has only structures where the cardinality is bounded to say it is not axiomatizable. I have lso tried to find a way to prove that is not axiomatizeable by completeness theorem (Vollständigkeitssatz) and EF games. But I cannot find a way. I also cannot provide an axiom system.
This turns out to have a rather silly answer: every graph is an element of some non-finitely-axiomatizable class of graphs, and so your class of graphs is just the class of all graphs (hence axiomatized by $\{\forall x(x=x)\}$).
Here's one way to prove this. (Below I conflate graphs with their isomorphism types, so I'm only looking at isomorphism-closed sets of graphs.)
First, consider a finite graph $G$. Let $\mathcal{X}$ be any non-finitely-axiomatizable set of graphs. If $\mathcal{X}$ contains $G$, we're done. Otherwise, consider $\mathcal{Y}=\mathcal{X}\cup\{H: H\cong G\}$; I claim that $\mathcal{Y}$ is also non-finitely-axiomatizable. To see this, note that since $G$ is finite there is a single sentence $\gamma$ characterizing $G$ up to isomorphism. If $\mathcal{Y}$ were finitely axiomatizable as $Mod(\psi)$ then we would have $\mathcal{X}=Mod(\psi\wedge\neg\gamma)$, which we know isn't the case.
OK, now consider an infinite graph $G$. The set $\{H: H\cong G\}$ cannot be finitely axiomatizable (or indeed elementary at all), since every elementary class containing an infinite structure contains arbitrarily large structures.
So we're done.
We can try to avoid this by rephrasing the question as something like the following:
The first bulletpoint above would still apply (in particular, note that $\mathcal{Y}=Mod(\{\varphi\vee\psi: \varphi\in Th(\mathcal{X})\}$), but the second would not. Instead, per bof's comment below we have to be a little more careful (and less silly): we observe that the set of all infinite graphs itself is not finitely axiomatizable.
(That's a consequence of the compactness theorem: the complement of a finitely axiomatizable class is an elementary class, and the class of finite graphs can't be elementary by compactness.)