Can we equip the power set $P$ of any set $S$ with a binary operation such that $P$ becomes a group (with some restrictions)? This question is inspired by this one.  Please read my answer there to get better context.
Settings. Let $S$ be a (not necessarily finite) set, and $P$ the power set of $S$ (i.e., $P$ is the set of all subsets of $S$).  A binary operator $*:P\times P\to P$ is said to be elementary if it can be given in terms of the standard set operations: the union operator $\cup$, the intersection operator $\cap$, the set difference operator $\setminus$, the symmetric difference operator $\triangle$, and the complement operator $(\_)^\complement$.
Some Examples.  This operator $\star$ is considered an elementary binary operator:
$$A \star B:= \big((M\setminus A)\cup (B\cap N)\big)^{\complement}\triangle \Big(A\cup B^\complement\Big)\text{ for all }A,B\subseteq S\,,$$
where $M$ and $N$ are fixed subsets of $S$.  On the other hand, if $|S|=2$, then this operator $\bullet$ is not an elementary binary operator:
$$A\bullet B:=\left\{
\begin{array}{ll}
S&\text{if }A\subseteq B\,,\\
\emptyset&\text{otherwise}\,,
\end{array}\right.$$
where $A,B\subseteq S$ (a proof can be done by imitating this answer).

Question.  For which groups $G$ of order $2^{|S|}$ does there exist an elementary binary operator $*:P\times P\to P$ such that $(P,*)$ is a group isomorphic to $G$?  If the case where $S$ is an infinite set is too troublesome, then an answer in the case where $S$ is a finite set is very welcome.

Let $n:=|S|$.  Write $Z_k$ for the cyclic group of order $k$.
Trivial Answer.  When $G\cong Z_2^n$, then the binary operator $\triangle$ does the work.  My conjecture is that there are no other groups.
Known Result.  When $|S|=2$ and $G\cong Z_4$, then there does not exist such an elementary binary operator.
 A: Let us identify $P$ with $\{0,1\}^S$ in the obvious way.  Then an elementary operation is just one which is given by applying some binary operation $\{0,1\}\times\{0,1\}\to\{0,1\}$ coordinatewise.  Indeed, it is clear that every elementary operation must have this form (since all the basic Boolean operations have this form), and conversely it is a simple exercise in Boolean algebra to build every binary operation on $\{0,1\}$ out of the basic Boolean operations.
So, $P$ must just be a product of copies of some binary operation on $\{0,1\}$.  As long as $S$ is nonempty, this means $P$ will be a group iff the corresponding operation on $\{0,1\}$ makes it a group (and the case where $S$ is empty is trivial).  But there is only one group operation on $\{0,1\}$ up to isomorphism, so $P$ can only be isomorphic to $\mathbb{Z}_2^S$.  (In fact, there are only two possible group operations at all: the usual symmetric difference operation and symmetric difference conjugated by swapping $0$ and $1$, which in terms of sets is just the complement of the symmetric difference.)
