Like Qiaochu says, the key here is that the direct product of finitely many abelian groups functions as both the (categorical) product and the (categorical) coproduct in the category of abelian groups.
And before your eyes glaze over, what that means is that:
A homomorphism from an abelian group $A$ into a (finite) direct product $G_1\times G_2\times\cdots\times G_n$ of abelian groups is equivalent to a family $f_1,\ldots,f_n$ of homomorphisms $f_i\colon A\to G_i$; (in fact, this holds for arbitrary groups, and arbitrarily many direct factors, not just finitely many); and
A homomorphism from a finite direct product $G_1\times G_2\times\cdots\times G_n$ of abelian groups into an abelian group $B$ is equivalent to a family $g_1,\ldots,g_n$ of homomorphisms $g_i\colon G_i\to B$ (here we do need both finiteness and abelianness).
The first equivalence is easy: given a map $f\colon A\to G_1\times\cdots\times G_n$, the maps $f_i$ are just the compositions of $f$ with the canonical projections $\pi_i\colon G_1\times\cdots\times G_n\to G_i$; going the other way, given a family $f_1,\ldots,f_n$ of maps, you get the map $A\to G_1\times\cdots\times G_n$ by $f(a) = (f_1(a),f_2(a),\ldots,f_n(a))$.
For the second equivalence, given a homomorphism $g\colon G_1\times\cdots\times G_n\to B$, we define the maps $g_i\colon G_i\to B$ by restricting $g$ to the subgroup $\{0\}\times\cdots \times \{0\}\times B_i\times\{0\}\times\cdots\times\{0\}$. Conversely, given a family of homomorphisms $g_1,\ldots,g_n$, we construct the map $g$ by $g(x_1,\ldots,x_n) = g_1(x_1)+g_2(x_2)+\cdots+g_n(x_n)$; here, both the fact that the product has only finitely many factors and that the groups are abelian is important.
Now let $A=B=G_1\times\cdots\times G_n$. Then a homomorphism from $A$ to $B$ is equivalent, by 1, to a family of homomorphisms $\Phi_j\colon A\to G_j$. And by 2, each $\Phi_j$ is equivalent to a family of homomorphisms $\phi_{ij}\colon G_i\to G_j$. Thus, each homomorphism from $A$ to $B$ is equivalent to a family $\{\phi_{ij}\mid 1\leq i,j\leq n\}$, with $\phi_{ij}\colon G_i\to G_j$.
Now suppose you have two homomorphisms, $\Phi,\Psi\colon A\to B$, and you want to compose them. If $\Phi$ corresponds to $\{\phi_{ij}\}$ and $\Psi$ corresponds to $\{\psi_{ij}\}$, what does the composition correspond to in terms of maps $G_i\to G_j$?
If you trace the correspondence carefully, you should find that the induced map from $G_i$ to $G_j$ is precisely
$$\psi_{i1}\circ\phi_{1j} + \psi_{i2}\circ\phi_{2j}+\cdots+\psi_{in}\circ\phi_{nj},$$
so that if you arrange the families $\{\phi_{ij}\}$ and $\{\psi_{ij}\}$ into matrices, composition corresponds to matrix multiplication in the usual way (though because composition is not commutative, you have to be mindful of the order of the products.
Once you have that endomorphisms can be "coded" as matrices with composition corresponding to matrix multiplication, the fact that automorphisms correspond to invertible matrices follows immediately. However, actually writing down a formula is complicated, because these matrices have entries that don't commute with one another; even in simple cases, like trying to do something like $C_{p^{\alpha}}\times C_{p^{\beta}}$ with $\alpha\gt\beta$, writing down the inverse of an automorphism in terms of its entries turns into a computation with congruences that is difficult to write down as a formula. But never fret, you aren't asked for an explicit formula.
First to get clear about the set $G$: it is the set of all subsets of a set $A$. (So the elements of $G$ are sets.) And by definition, $G$ is therefore the powerset of $A$, denoted $G =\mathcal{P}(A))$. So $|G| = 2^{|A|}$, which is finite if and only if $|A|$ is finite, and is infinite otherwise.
The operation on the sets $g_1, g_2 \in G$ is the symmetric difference of $\;g_1\;$ and $\;g_2\;$ which we'll denote as $\;g_1\;\triangle\;g_2\;$ and is defined as the set of elements which are in either of the sets but not in their intersection: $\;g_1\;\triangle\;g_2 \;= \;(g_1\cup g_2) - (g_1 \cap g_2)\tag{1}$
Hence, the symmetric difference $g_i \;\triangle\; g_j \in G\;$ for all $\;g_i, g_j \in G$. ($G$ is the set of ALL subsets of $A$, so it must include any possible set resulting from symmetric difference between any arbitrary sets in $G$, which are also subsets of $A$). That is we have now established, that the symmetric difference is closed on $G$.
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 $g_i \in G, g_i\;\triangle \; g_i = \varnothing$).
The justifications for these properties is very straightforward, but good to include for a proof that the symmetric difference, together with the set $G$ as defined, form an abelian group.
So, you've covered most of the bases, but you simply want to confirm/note the closure of the symmetric difference operation on $G$ and why, add a bit of justification for the identity and inverse claims, and to address whether, or when, $G$ is finite/infinite: this last point being that the order of $G$ depends on the cardinality of $A$.
Best Answer
I believe the "disjoint union" alluded to by David Holden means "disjunctive union" which is a synonym for the symmetric difference. So, you aren't listing multiple ways, really.
As the wiki clearly states: The power set of any set becomes a Boolean ring with symmetric difference as the addition of the ring and intersection as the multiplication of the ring.
This is the standard ring structure on the powerset of a set. It may have other ring structures, but none would arise so naturally as this one, and would probably not have meaningful connection to the fact the underlying set is a powerset.