Theorem. The Kunen inconsistency works over ZFC with atoms. That is, in this theory, there is no non-identity elementary embedding $j:V\to V$ that fixes every atom.
Proof. Suppose that $j:V\to V$ is an elementary embedding fixing every atom. Let $W$ be the atomless part of $V$, the sets that have no atoms in their transitive closures. It follows that $W$ is a model of the usual ZFC, without atoms. Since $W$ is definable, it follows that $j\upharpoonright W$ is an elementary embedding from $W$ to $W$. I claim that it is nontrivial, which will contradict Kunen's theorem.
So suppose $j$ is trivial on $W$, and in particular, $j$ fixes every ordinal.
For any set of atoms $A$, define the rank hierarchy over $A$ by $V_0(A)=A$, $V_{\alpha+1}(A)=P(V_\alpha(A)$ and $V_\lambda(A)=\bigcup_{\alpha<\lambda}V_\alpha(A)$ at limits.
Since $j(A)$ is a set of atoms, but can contain only the atoms in $A$, it follows that $j(A)=A$. Thus, since $j$ also fixes ordinals, it follows that $j(V_\alpha(A))=V_\alpha(A)$. (This observation seems to address the concern you mentioned in your question.)
An inductive argument now shows that $j$ is the identity on the elements of every $V_\alpha(A)$. If this is true at $\alpha$, then it is true for the elements of $V_{\alpha+1}(A)$, and the statement carries trivially through limits.
This implies $j(u)=u$ for every set $u$, since every $u$ is in $V_\alpha(A)$ for the set $A$ of atoms appearing in its transitive closure. $\Box$