Proving that set $(-1,1)$ with the operation $a*b=\frac{x+y}{1+xy}$ is an abelian group. I am trying to prove that $(G, *)$ is an abelian group with $G=(-1,1)$ and $a*b=$$\frac{x+y}{1+xy}$.
Thus far I have found that the identity element $e=0$. From here, I set $a*b=0$ and found $a^{-1}$ to be $-a$. 
My work for trying to prove closure and that the set is abelian is:
Let $a,b \in G$, where $a=e$ and $b=a^{-1}$. Does  $a*b=b*a$?
Evaluating, I see that $a^{-1}=a^{-1}$ which then proves that G is closed and G is an abelian group.
Did I go about this proof correctly and efficiently? 

You have found neutral and inverse fine.
You need to show closure, i.e. if $-1<x,y<1$, is then also $-1<\frac{x+y}{1+xy}<1$? (To really be precise: Is the expression defined for all cases in the first place?)
Abelian is quite clear, because $x+y=y+x$ and $xy=yx$.
However, the toughest part (at least if one is forced to do direct computatons) is in fact associativity: This boils down to writing down the complicated fractions for $x*(y*z)$ and $(x*y)*z$ and showing that they are indeed equal.
