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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 $2$).

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 $n$.

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 $2$).

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 $$((x\oplus x)\oplus(x\oplus x))_{n+2}=x_n.\tag1$$ 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 $$(x\oplus(x\oplus(x\oplus x)))_{n+2}=x_{n+1}x_nx_{n-1}.\tag2$$ If $(x\oplus x)\oplus(x\oplus x)=x\oplus(x\oplus(x\oplus x))$, then from $(1)$ and $(2)$ we have $$x_n=((x\oplus x)\oplus(x\oplus x))_{n+2}=$$ $$(x\oplus(x\oplus(x\oplus x)))_{n+2}=x_{n+1}x_nx_{n-1},$$ whence $x_n=1\implies x_{n+1}=x_{n-1}=1$, i.e., $x$ is a constant function. Since $x_n=0$ for $n\lt0$, it follows that $x_n=0$ for all $n$.

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 $$x_n=((x\oplus x)\oplus(x\oplus x))_{n+2}=(x\oplus(x\oplus(x\oplus x)))_{n+2}=0.$$

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 $2$).

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 $n$.

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 $$x_n=((x\oplus x)\oplus(x\oplus x))_{n+2}=(x\oplus(x\oplus(x\oplus x)))_{n+2}=0.$$

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 $$x_n=((x\oplus x)\oplus(x\oplus x))_{n+2}=(x\oplus(x\oplus(x\oplus x)))_{n+2}=0.$$