Following the suggestion in a comment by [Achim Krause](https://mathoverflow.net/users/39747/achim-krause) I will show that the only solution of the equation $(x\oplus x)\oplus(x\oplus x)=x\oplus(x\oplus(x\oplus x))$ in the magma $(\mathcal P(\mathbb N),\oplus)$ is $x=\emptyset$, whence $\{\emptyset\}$ is the only associative submagma of $(\mathcal P(\mathbb N),\oplus)$.

It will be convenient to identify an element of $\mathcal P(\mathbb N)$ with its characteristic function as a subset of $\mathbb Z$, so that for $x\in\mathcal P(\mathbb N)$ and $n\in\mathbb Z$ we have $x_n=1$ if $n\in x$ and $x_n=0$ if $n\notin x$; in particular $x_n=0$ for all $n\lt0$. Thus for all $n\in\mathbb Z$ and $x,y\in\mathcal P(\mathbb N)$ we have $(x\oplus y)_n=x_n+y_n+x_{n-1}y_{n-1}$ (addition modulo $2$).

Now, for any $x\in\mathcal P(\mathbb N)$ and any $n\in\mathbb Z$, we have
$$(x\oplus x)_n=x_n+x_n+x_{n-1}x_{n-1}=x_{n-1}$$
and
$$((x\oplus x)\oplus(x\oplus x))_n=(x\oplus x)_{n-1}=x_{n-2},$$
so that
$$((x\oplus x)\oplus(x\oplus x))_{n+3}=x_{n+1}.\tag1$$
Also
$$(x\oplus(x\oplus x))_n=x_n+x_{n-1}+x_{n-1}x_{n-2}$$
and
$$(x\oplus(x\oplus(x\oplus x)))_n=$$$$x_n+(x_n+x_{n-1}+x_{n-1}x_{n-2})+x_{n-1}(x_{n-1}+x_{n-2}+x_{n-2}x_{n-3})$$$$=x_{n-1}x_{n-2}x_{n-3},$$
so that
$$(x\oplus(x\oplus(x\oplus x)))_{n+3}=x_{n+2}x_{n+1}x_n.\tag2$$
If $(x\oplus x)\oplus(x\oplus x)=x\oplus(x\oplus(x\oplus x))$, then from $(1)$ and $(2)$ we have
$$x_{n+1}=((x\oplus x)\oplus(x\oplus x))_{n+3}=$$$$(x\oplus(x\oplus(x\oplus x)))_{n+3}=x_{n+2}x_{n+1}x_n,$$
whence $x_n=0\implies x_{n+1}=0$. Since $x_n=0$ for $n\lt0$, it follows that $x_n=0$ for all $n$.