It's not clear to me exactly what structure you're looking at, but I suspect it's (equivalent to) the $\{\in\}$-structure $\mathfrak{S}=M\sqcup P(M)$. (If not, what do you mean?) Under this interpretation the answer to your question is no. Consider $M$ as an element of $\mathfrak{S}$. $M$ is certainly a definable subclass of $M$, so we must have $j$ send $M$ to some $a\in M$; in particular, $j(M)\not=M$ (where again we're thinking of $M$ here as an element of our "power structure" $\mathfrak{S}$, not as a subset of it).

EDIT: Based on the comments, I think it's worth clarifying a bit of the nature of the Boffa-Jensen construction of a model of NFU (which appears to be part of the motivation for this question).

In this construction, we do not have an elementary embedding; rather, we begin with a model $M$ of ZFC and an automorphism - unfortunately for our context denoted "$j$" in Holmes' article, but which I will call "$a$" - which moves a rank, that is, which is not the identity on the ordinals. I think the conflation of automorphism and elementary embedding is generating some confusion here.

Incidentally, a fun exercise is to show that any nontrivial automorphism of a model of ZFC must move an ordinal, while this is not true of ZF! I believe this was first observed by Harvey Friedman. A cute result of Cohen is the construction of a (necessarily ill-founded) model of ZF with a nontrivial automorphism of finite order, and it was in Cohen's paper that I first saw this fact quoted.

Now we come to a key point: the automorphism $a$ is completely external, and the mere existence of a nontrivial automorphism immediately implies that $M$ is illfounded. Put better: any well-founded model of ZFC, or ZF, or even much weaker, has no nontrivial automorphisms. (Think about the least rank of an element moved by a hypothetical nontrivial automorphism …) Indeed, when we speak of automorphisms of models of set theory these are always understood as being completely external to the models in question; note that this is in contrast with the situation for elementary embeddings, where there are various flavors of definability. Similarly, the connection with ill-foundedness breaks down for elementary embeddings: assuming $0^\sharp$ exists, there is a nontrivial elementary embedding of $L$ into itself, but $L$ is of course well-founded. Automorphisms and elementary embeddings are really very different objects.

Anyways, in the Boffa-Jensen construction we take an $M$ and $a$ as above, and fix some $M$-ordinal $\alpha$ with $a(\alpha)\not=\alpha$. WLOG we have $a(\alpha)<\alpha$ (otherwise replace $a$ with $a^{-1}$. The structure $\mathcal{B}$ associated to the data $(M, a, \alpha)$ has domain $(V_\alpha)^M$, and elementhood relation $\varepsilon$ given by $$x\varepsilon y\iff a(x)\in^My\wedge y\in^MV_{a(\alpha)+1}.$$ Everything in $V_\alpha^M\setminus V_{a(\alpha)+1}^M$ winds up being an urelement in the sense of $\mathcal{B}$: if $y\in V_\alpha^M\setminus V_{a(\alpha)+1}^M$, then we never have $x\varepsilon y$ for any $x\in V_\alpha^M$.

(Why is $V_\alpha^M\setminus V_{a(\alpha)+1}^M$ nonempty? Well, one way to guarantee this would be to make sure that $M$ is an $\omega$-model, that is, the "finite ordinals" in the sense of $M$ are in fact all finite; however, even this assumption is unnecessary, since $a(\alpha)$ and $\alpha$ must have the same "$M$-parity" and no ordinal has the same parity as its successor.)

However, we're seeing something very specific to automorphisms here. An elementary embedding $M\preccurlyeq M$ simply wouldn't help us do the same thing: it is crucial that $a$ moves ranks downwards. Having an "upward-moving" automorphism is still fine, since we can invert it, but elementary embeddings aren't surjective in general so no similar trick will work there. Indeed, I know of no way to get a model of NFU (or similar) from, for example, $0^\sharp$ in a natural way (remember that $0^\sharp$ is equivalent in a sense to a nontrivial elementary embedding from $L$ into itself; this isn't a contradiction with Kunen, since "$0^\sharp$ exists" implies $V\not=L$).

It's not clear to me exactly what structure you're looking at, but I suspect it's (equivalent to) the $\{\in\}$-structure $\mathfrak{S}=M\sqcup P(M)$. (If not, what do you mean?) Under this interpretation the answer to your question is no. Consider $M$ as an element of $\mathfrak{S}$. $M$ is certainly a definable subclass of $M$, so we must have $j$ send $M$ to some $a\in M$; in particular, $j(M)\not=M$ (where again we're thinking of $M$ here as an element of our "power structure" $\mathfrak{S}$, not as a subset of it).

It's not clear to me exactly what structure you're looking at, but I suspect it's (equivalent to) the $\{\in\}$-structure $\mathfrak{S}=M\sqcup P(M)$. (If not, what do you mean?) Under this interpretation the answer to your question is no. Consider $M$ as an element of $\mathfrak{S}$. $M$ is certainly a definable subclass of $M$, so we must have $j$ send $M$ to some $a\in M$; in particular, $j(M)\not=M$ (where again we're thinking of $M$ here as an element of our "power structure" $\mathfrak{S}$, not as a subset of it).

But $M$ is a definable element of $\mathfrak{S}$: $M$ is the unique $z\in\mathfrak{S}$ such that for all $a\in\mathfrak{S}$ we have $\mathfrak{A}\models\exists x(a\in x)\iff a\in z$. So $M$ can't be moved by an elementary embedding.

As far as I can tell, this argument (or a version of it) works for any reasonable envisioning of $P(M)$ as a structure.

It's not clear to me exactly what structure you're looking at, but I suspect it's (equivalent to) the $\{\in\}$-structure $\mathfrak{S}=M\sqcup P(M)$. (If not, what do you mean?) Under this interpretation the answer to your question is no. Consider $M$ as an element of $\mathfrak{S}$. $M$ is certainly a definable subclass of $M$, so we must have $j$ send $M$ to some $a\in M$; in particular, $j(M)\not=M$ (where again we're thinking of $M$ here as an element of our "power structure" $\mathfrak{S}$, not as a subset of it).

But $M$ is a definable element of $\mathfrak{S}$: $M$ is the unique $z\in\mathfrak{S}$ such that for all $a\in\mathfrak{S}$ we have $\mathfrak{A}\models\exists x(a\in x)\iff a\in z$. So $M$ can't be moved by an elementary embedding.

As far as I can tell, this argument (or a version of it) works for any reasonable envisioning of $P(M)$ as a structure. For example, another possible interpretation of your question eschews "$\in$" but has the same result. We look at a structure $\mathfrak{S}$ whose elements correspond to subsets of $M$ (if you want, the domain of $\mathfrak{S}$ is exactly $P(M)$). The language of $\mathfrak{S}$ is just $\{\subseteq\}$, interpreted in the obvious way.

The domain of $\mathfrak{S}$ has a special subset $Ground$: namely, the set of all $a$ with $\{x\in M: x\in a\}=\{x\in M: M\models x\in b\}$ for some $b\in M$. These are the "internal sets" of $M$.

The first two requirements you put on $j$ are elementarity and $j(a)\in D$ whenever $a$ is a definable subset of $M$. The latter means in particular that we must have $j(M)\in D$ (note that $M\in \mathfrak{S}$). But then $j$ certainly can't be elementary, since $M$ is the unique $a\in\mathfrak{S}$ satisfying $\forall x(x\subseteq M)$ (and elementary maps can't move definable elements.