Is it ok to prove a subset of a group is an abelian group this way? I'll admit from the start I'm being lazy, but all the same it makes thing's neater in my opinion - if it's valid.
Now it's known that if we have a group $G$ such that $g^2=e,\ \ \forall g \in G$ then $G$ is abelian. But what if we don't know G is a group yet? We know the mentioned condition, but thats it. Is it ok to just check, say $x,y \in G \implies xy \in G$ and take the other order for granted? Since.. well, the group, if it is one, will be abelian!
In particular, I'd like to show that $G=\{e,(12)(34),(13)(24),(14)(23)\}\subset S_4$ is an abelian group.
 A: As every element has order two you do not need to check both directions of every multiplication, i.e., you have $3$ non-identity elements and you only need to check $\binom32 = 3$ multiplications instead of the $6$ that a brute force approach would normally require.
The reason is because if $xy = z$ and $z$ is in your subset, then because $z$ has order two we know that $z^{-1} = z$ is in your subset, and $z^{-1} = (xy)^{-1} = y^{-1}x^{-1} = yx$ (because $x$ and $y$ also have order $2$).
That trick I just used is exactly how you prove that if every element of a group is order $2$ it is abelian, but I don't want to say that it's because of that theorem that you don't have to check the other multiplications.  Really it's because of the proof of that theorem; because the method of proof still applies in this circumstance.  I make this distinction because if you have a theorem that says "If a group satisfies condition $P$ then that group is a $Q$" and then you check that a subset satisfies condition $P$, then in general it is NOT true that this implies that your subset is in fact a subgroup of type $Q$.
A: When taking arbitrary elements $x, y \in G$ to test for closure (to test whether $x, y \in G \implies xy \in G$), then the order of the product doesn't matter. 
That is, if you can show $xy \in G$ in that situation, then you've proven closure of the candidate group under the group operation regardless of whether the candidate group is abelian or not. So it is never necessary to show, additionally, that $yx \in G$ when trying to establish that $G$ is closed.
UPDATE: Given your clarification about the specific set in question:
As you note, we know only that a subgroup (group) with every element of order two is abelian. In order to conclude that your set is therefore abelian, we need first to establish that the proposed set is a subgroup (group) of $S_4$. Then once that's done, you can conclude it must be abelian, based on your knowledge of groups whose elements are all of order $2$. 
