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=$$\frac{x+y}{1+xy}$.
Thus far I have found that the identity element $e=0$. From here, I set $a*b=0$ and found $a^{-1}$ to be $-a$. 
My work for trying to prove closure and that the set is abelian is:
Let $a,b \in G$, where $a=e$ and $b=a^{-1}$. Does  $a*b=b*a$?
Evaluating, I see that $a^{-1}=a^{-1}$ which then proves that G is closed and G is an abelian group.
Did I go about this proof correctly and efficiently? 
 A: You have found neutral and inverse fine.
You need to show closure, i.e. if $-1<x,y<1$, is then also $-1<\frac{x+y}{1+xy}<1$? (To really be precise: Is the expression defined for all cases in the first place?)
Abelian is quite clear, because $x+y=y+x$ and $xy=yx$.
However, the toughest part (at least if one is forced to do direct computatons) is in fact associativity: This boils down to writing down the complicated fractions for $x*(y*z)$ and $(x*y)*z$ and showing that they are indeed equal.
A: If $|a|<1$ and $|b|<1$, then also $|ab|<1$, so $-1<ab<1$ and $0<1+ab<2$. Consequently, $(a+b)/(1+ab)$ is defined for all $a,b\in (-1,1)$. Now, for $a,b\in(-1,1)$,
$$
\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
$$
for all $a,b\in G$.
A: Another way to look at this is as follows. Note that 
$$tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}}$$ defines a bijective map from $\mathbb{R} \rightarrow (-1,1).$ And $$tanh(a+b)=\frac{tanh(a)+tanh(b)}{1+tanh(a)tanh(b)}.$$ And obviously addition carries over in the operation $*$ defined by the OP.
A: The identity is indeed 0 and $a^{-1}$ is indeed $-a$. You have 3 things left to show:
Closure: You need to show that if a and b are in (-1, 1) then so is $\frac{a+b}{1+ab}$. It seems like the best way to do this would be to consider separate cases when a and b are positive or negative. 
Associativity: You need to prove $a*(b*c) = (a*b)*c$. This should be straightforward, albeit messy algebra.
Commutativity: What you have done is show that the identity element commutes with group elements; you need to show an arbitrary group elements commutes with other group elements. This is clear, since $\forall a,b \in G$,   $$a*b = \frac{a+b}{1 + ab} = \frac{b+a}{1 + ba} = b*a $$
