Following the suggestion in a comment by 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 $2$) and $(x\oplus x)_n=x_{n-1}$.
Now suppose $x\in\mathcal P(\mathbb N)$ and $(x\oplus x)\oplus(x\oplus x)=x\oplus(x\oplus(x\oplus x))$; I claim that $x_n=0$ for all $n\in\mathbb Z$. It will suffice to show that $x_n=0$ assuming that $x_i=0$ for all $i\lt n$. On the one hand we have $$((x\oplus x)\oplus(x\oplus x))_{n+2}=(x\oplus x)_{n+1}=x_n.\tag1$$ On the other hand: $$(x\oplus(x\oplus x))_n=x_n+(x\oplus x)_n+x_{n-1}(x\oplus x)_{n-1}=x_n+0+0=x_n;$$ $$(x\oplus(x\oplus x))_{n+1}=x_{n+1}+(x\oplus x)_{n+1}+x_n(x\oplus x)_n=x_{n+1}+x_n;$$ $$(x\oplus(x\oplus x))_{n+2}=x_{n+2}+(x\oplus x)_{n+2}+x_{n+1}(x\oplus x)_{n+1}$$$$=x_{n+2}+x_{n+1}+x_{n+1}x_n;$$ $$(x\oplus(x\oplus(x\oplus x)))_n=x_n+(x\oplus(x\oplus x))_n+x_{n-1}(x\oplus(x\oplus x))_{n-1}$$$$=x_n+x_n+0=0;$$ $$(x\oplus(x\oplus(x\oplus x)))_{n+1}=x_{n+1}+(x\oplus(x\oplus x))_{n+1}+x_n(x\oplus(x\oplus x))_n$$$$=x_{n+1}+(x_{n+1}+x_n)+x_n=0;$$ and finally $$(x\oplus(x\oplus(x\oplus x)))_{n+2}=$$$$x_{n+2}+(x\oplus(x\oplus x))_{n+2}+x_{n+1}(x\oplus(x\oplus x))_{n+1}=$$$$x_{n+2}+(x_{n+2}+x_{n+1}+x_{n+1}x_n)+x_{n+1}(x_{n+1}+x_n)=0.\tag2$$ From $(1)$ and $(2)$ we have $$x_n=((x\oplus x)\oplus(x\oplus x))_{n+2}=(x\oplus(x\oplus(x\oplus x)))_{n+2}=0.$$