Binary operations do not need to be surjective. Here is a natural example:
Let $\mathbb N = \{1,2,3,\dots \}$. Then $+: \mathbb N \times \mathbb N \to \mathbb N$ is not surjective because $1$ is not in the image.
Here is another natural, more interesting example:
Let $\mathbb N' = \{2,3,\dots \}$. Then $\times: \mathbb N' \times \mathbb N' \to \mathbb N'$ is not surjective because the prime numbers are not in the image.
Let $(B,+,\cdot)$ be the Boolean algebra with two generators $u$ and $v$. The multiplication in $B$ is given by $u\cdot u=u$, $v\cdot v=v$, and $u\cdot v=v\cdot u=0$. Therefore, $e:=u+v$ is the multiplicative identity of $B$.
We identify $0$, $u$, $v$, and $e$ with $\emptyset$, $\{a\}$, $\{b\}$, and $\{a,b\}$, respectively. Then, we can associate any set operation on $\mathcal{P}(S)$ with a polynomial operator in $B$. This is because the symmetric difference operator $\triangle$ is associated to the polynomial $d(x,y):=x+y$, the union operator $\cup$ is associated to the polynomial $f(x,y):=x+y+x\cdot y$, the intersection operator $\cap$ is associated to the polynomial $g(x,y):=x\cdot y$, the set difference operator $\setminus$ is associated to $h(x,y):=x+x\cdot y$, and the complement operator is associate to the polynomial $k(x):=e+x$.
Suppose that there exists a polynomial $p(x,y)\in B[x,y]$ such that the binary operation on $\mathcal{P}(S)$ equips $\mathcal{P}(S)$ with a structure of $G:=\mathbb{Z}/4\mathbb{Z}$. Let $z\in B$ be the element that acts as the identity of $G$. Since $G$ is abelian, we get $p(x,y)=p(y,x)$, whence
$$p(x,y)=\alpha+\beta\cdot x+\beta\cdot y+\gamma\cdot x\cdot y$$
for some $\alpha,\beta,\gamma\in B$. Now,
$$0=p(0,z)=\alpha+\beta\cdot z\,.$$
Therefore,
$$\beta\cdot z=\alpha\,.$$
We also have
$$z=p(z,z)=\alpha+\beta\cdot z+\beta\cdot z+\gamma\cdot z\cdot z=\alpha+\gamma\cdot z\,.$$
Hence,
$$(e+\gamma)\cdot z=z+\gamma\cdot z=\alpha\,.$$
Furthermore,
$$\begin{align}e=p(e,z)&=\alpha+\beta\cdot e+\beta\cdot z+\gamma\cdot e\cdot z
\\&=\alpha+\beta+\alpha+(\alpha+z)=\alpha+\beta+z\,.\end{align}$$
Consequently,
$$z=e+\alpha+\beta\,.$$
From $\beta\cdot z=\alpha$, we conclude that $\alpha\cdot\beta=\alpha$, or $$\alpha\cdot(e+\beta)=0\,.$$
Case I: $\beta=0$. Then, $\alpha=\beta\cdot z=0$. Therefore, $z=e+\alpha+\beta=e$. As $(e+\gamma)\cdot z=\alpha$, we conclude that $\gamma=e$. Hence, $p(x,y)=x\cdot y$, which clearly does not work. (Alternatively, note that $p(0,0)=0$, which contradicts the result that $z=e$ is the identity of $G$.)
Case II: $\beta=u$. Then, $\alpha\cdot v=\alpha\cdot(e+\beta)=0$. Hence, either $\alpha=0$ or $\alpha=u$.
If $\alpha=0$, then from $z=e+\alpha+\beta$, we get $z=v$. From $(e+\gamma)\cdot z=\alpha$, we conclude that $\gamma=0$ or $\gamma=v$. In the case $\gamma=0$, we get $p(x,y)=u\cdot(x+y)$, which means that the image of $p(x,y)$ can only be $0$ or $u$, leading to a contradiction. In the case $\gamma=v$, we get $$p(x,y)=u\cdot(x+y)+v\cdot(x\cdot y)\,,$$
whence
$$p(u,0)=u\cdot(u+0)+v\cdot(u\cdot 0)=u\,,$$
but this contradicts the conclusion that $z=v$ is associated to the identity of $G$.
If $\alpha=u$, then $z=e+\alpha+\beta=e$. From $(e+\gamma)\cdot z=\alpha$, we conclude that $\gamma=v$. Ergo,
$$p(x,y)=u+u\cdot(x+y)+v\cdot(x\cdot y)\,.$$
Thus,
$$p(u,u)=u+u\cdot(u+u)+v\cdot(u\cdot u)=u\,.$$
This contradicts the result that $z=e$ is associated to the identity of $G$.
Case III: $\beta=v$. The argument is the same as Case II.
Case IV: $\beta=e$. Then, $z=e+\alpha+\beta=\alpha$, and from $(e+\gamma)\cdot z=\alpha$, we get $\gamma\cdot\alpha=0$.
If $\alpha=0$, then $z=0$ and $$p(x,y)=(x+y)+\gamma\cdot(x\cdot y)\,.$$
Therefore, $p(\gamma,\gamma)=\gamma$ implies that $\gamma$ is associated to the identity of $G$, making $\gamma=z=0$. Thus, $p(x,y)=x+y$, which clearly does not work. (Alternatively, note that $p(0,0)=0$, which contradicts the result that $z=e$ is the identity of $G$.)
If $\alpha=u$, then $z=u$ and $$p(x,y)=u+(x+y)+\gamma\cdot(x\cdot y)\,.$$ Note that $\gamma\cdot \alpha=0$ implies $\gamma=0$ or $\gamma=v$. If $\gamma=0$, then $p(0,0)=u=p(v,v)$, which contradicts the fact that $G$ has only one element of order $2$. If $\gamma=v$, then $p(e,v)=v$, which contradicts the result that $u$ is associated to the identity of $G$.
If $\alpha=v$, then we have a similar contradiction to the previous subcase.
If $\alpha=e$, then $z=e$ and $\gamma=0$, making $$p(x,y)=e+(x+y)\,.$$ Now, $p(x,x)=e$ for all $x\in B$ contradicts the fact that $G$ has only one element of order $2$.
Therefore, such a polynomial $p(x,y)\in B[x,y]$ does not exist. Hence, there is no binary operator $*$ on $\mathcal{P}(S)$ given by the usual set operations that makes $\mathcal{P}(S)$ isomorphic to the group $\mathbb{Z}/4\mathbb{Z}$.
P.S. See a much simpler argument to a more generalized setting here.
Best Answer
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.)