I thought it might be helpful to add a completely different viewpoint. (Feel free to ignore this answer, if other answers are clearer for you or if this post contains some things that you have not studied yet.)
We work with subsets of some given set $M$. Such subsets can be identified with function from $M$ to $\{0,1\}$, namely, there is a bijection between $\mathcal P(M)$ and $\{0,1\}^M$ given by $A\mapsto\chi_A$ where $\chi_A$ is the characteristic function of the set $A$:
$\chi_A(x)= \begin{cases} 1,& x\in A\\ 0,& x\notin A \end{cases} $
So we have a bijection between $\mathcal P(M)$ and the ring $\mathbb Z_2^M$ (with the componentwise operations). Let us have a look which set operations correspond to the ring operations $\oplus$ and $\odot$.
$\chi_{A\triangle B}=\chi_A\oplus\chi_B$ (I prefer the symbol $\triangle$ as the notation for symmetric difference.) $\chi_{A\cap B}=\chi_A\odot\chi_B$
So $(\mathcal P(M),\triangle,\cap)$ is a ring isomorphic to $(\mathbb Z_2^M, \oplus, \odot)$. If we work with the second ring, we see that the identity for the operation $\oplus$ is $(0,0,\dots,0)$, which means that the identity for $\triangle$ is the corresponding set, which is $\emptyset$.
I am mentioning this basically for the reason that it illustrates the situation when instead of proving some things about one structure (in this case it is $(\mathcal P(M),\triangle,\cap)$) it might be helpful to find an isomorphic structure which we are already familiar with.