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.

So we've established that $j$ must move ordinals, and so $j\upharpoonright W:W\to W$ is a nontrivial elementary embedding of $W$. By restricting to $V_{\lambda+2}^W$, where $\lambda$ is the supremum of the critial sequence, we get a set embedding $j\upharpoonright V_{\lambda+2}:V_{\lambda+2}\to V_{\lambda+2}$, which is an element of $W$, and this contradicts the Kunen inconsistency in $W$.\Box$

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. If $j$ is not the identity embedding, then I claim that $j$ must move an ordinal. To see this, let $u$ be any $\in$-minimal element that is moved by $j$. Let us well-order $u$ in some order type $\beta$. Since ordinals are fixed by $j$, it follows that the $\alpha^{th}$ element of $u$ is carried by $j$ to the $j(\alpha)^{th}=\alpha^{th}$ element of $j(u)$, but since those elements are fixed (by minimality of $u$) and the total length of the well-order is fixed, it follows that $j(u)$ is simply $u$ itself, contradicting the choice of $u$. (In fact, by an argument on ranks, one can eliminate the use of choice in this part of the argument; but you still need it for the other part.)

So $j$ is not the identity on ordinals. We may now restrict $j$ to the pure part of the universe. Let $W$ be the class of sets having no atoms in their transitive closures. This is a model of ZFC, without atoms. Since this is definable, it follows that $j\upharpoonright W$ is a nontrivial elementary embedding from $W$ to $W$. This contradicts the usual Kunen inconsistency. For example, one gets that $j\upharpoonright V_{\lambda+2}^W$ is a set in $W$, where $\lambda$ is the supremum of the critical sequence, and this violates the usual ZFC version of the Kunen inconsistency in $W$. $\Box$

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. $\Box$

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.

So we've established that $j$ must move ordinals, and so $j\upharpoonright W:W\to W$ is a nontrivial elementary embedding of $W$. By restricting to $V_{\lambda+2}^W$, where $\lambda$ is the supremum of the critial sequence, we get a set embedding $j\upharpoonright V_{\lambda+2}:V_{\lambda+2}\to V_{\lambda+2}$, which is an element of $W$, and this contradicts the Kunen inconsistency in $W$.\Box$