You're not mistaken. You just proved that there is no "set of all graphs". In fact, there is no set of all singletons either. And so definitely there will be no set of all graphs.
In some set theories which admit a universal set (e.g. Quine's New Foundations), this is still not a problem, because the function mapping a set on its power set is not itself a set, so there is no contradiction to Cantor's theorem. (The reason this is happening is that the existence of a universal set means that not every formula can be used to define a subset.)
(Also, one minor mistake you have here, is that you didn't actually prove that $G=\mathcal P(G)$, but rather that there is a surjection from $G$ onto its power set. Note that a graph is not just a set, it's an ordered pair of set [of vertices] and a set of edges. So it isn't usually the case that a set of ordered pairs is itself an ordered pair.)