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? 

If $|a|<1$ and $|b|<1$, then also $|ab|<1$, so $-1<ab<1$ and $0<1+ab<2$. Consequently, $(a+b)/(1+ab)$ is defined for all $a,b\in (-1,1)$. Now, for $a,b\in(-1,1)$,
$$
\frac{a+b}{1+ab}<1
$$
becomes $a+b<1+ab$ or $1-a-b+ab>0$, that is, $(1-a)(1-b)>0$, which is true. Similarly
$$
\frac{a+b}{1+ab}>-1
$$
becomes $a+b>-1-ab$ or $(1+a)(1+b)>0$. Again this is true, under our hypotheses.
Let's tackle $a*(b*c)$; since $b*c=(b+c)/(1+bc)$, we have
$$
a*(b*c)=\frac{a+\dfrac{b+c}{1+bc}}{1+a\dfrac{b+c}{1+bc}}=
\frac{abc+a+b+c}{1+bc+ab+ac}
$$
Can you try with $(a*b)*c$?
You computed the identity and the inverse correctly. However, being $G$ abelian means
$$
a*b=b*a
$$
for all $a,b\in G$.
