It is not that associativity is required for groups... That is quite backwards: the truth is actually that groups are associative.
Your question seems to come from the idea that people decided how to define groups and then began to study them and find them interesting. In reality, it happened the other way around: people had studied groups way before actually someone gave a definition. When a definition was agreed upon, people looked at the groups they had at hand and saw that they happened to be associative (and that that was a useful piece of information about them when working with them) so that got included in the definition.
If I may say so, it is this which is important to understand. The way we teach abstract algebra nowdays somewhat obscures this fact, but this is how essentially everything comes to be.
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
The associativity property is an algebraic identity that the group operation has to satisfy: $(ab)c=a(bc)$. Whether this identity is true for three fixed elements $a$, $b$, and $c$ does not depend on what set I put them in.
The other properties you mentioned are fundamentally different from associativity because they assert the existence of an object in the subgroup: