Let $A$ be a non empty set. Let $P(A)$ denote the power set of $A$.
$P(A)$ can be given a group structure in multiple ways.
1. Using Disjoint union a group operation Source
may be worth noting in this context that the power set of $X$ is a group under the operation of disjoint union, with the empty set as identity. if the complement of $A \subset X$ is $A'$ then the disjoint union of $A$ and $B$ is: $$ A \circ B = (A \cap B') \cup (A' \cap B) $$ proving associativity is a nice exercise, though it follows much more simply if the disjoint union is viewed as ordinary addition (mod 2) in $\mathbb{F}_2^X$
2. Using Symmetric Difference as Group Operation Source
The power set $G$ of any set $A$ becomes an abelian group under the operation of symmetric difference:
Why abelian? Easy to justify, just use the definition in $(1)$ above: it's defined in a way that $g_1 \triangle g_2$ means exactly the same set as $g_2 \triangle g_1$, for any two $g \in G$.
As you note, the symmetric difference on $G$ is associative, which can be shown using the definition in $(1)$, by showing for any $f, g, > h \in G, (f\; \triangle\; g) \triangle \;h = f\;\triangle\; (g > \;\triangle\; h)$.
The empty set is the identity of the group (it would be good to justify this this, too), and
every element in this group is its own inverse. (Can you justify this, as well? Just show for any $g_i \in G, g_i\;\triangle \; g_i = > \varnothing$).
Question: Can we define a ring structure on power set of $A$.