The group law is given by $a \oplus b = \frac{a+b}{1+a\,b}$ for $a,b \in G = (-1,1)$
For associativity I guess I have to show $(a \oplus b) \oplus c = a \oplus (b \oplus c)$, hence:
$$\begin{equation}\frac{\frac{a+b}{1+a\,b}+c}{1+\frac{a+b}{1+a\,b}\,c}= \frac{\frac{c+b}{1+c\,b}+a}{1+\frac{c+b}{1+c\,b}\,a}\end{equation}$$
but I got a hard time doing so. Is this actually the task?
On
We have
$$\begin{align} (a\oplus b)\oplus c&=\left(\frac{a+b}{1+ab}\right)\oplus c\\ &=\frac{\left(\frac{a+b}{1+ab}\right)+c}{1+\left(\frac{a+b}{1+ab}\right)c}\\ &=\frac{(a+b)+c(1+ab)}{(1+ab)+(a+b)c}\\ &=\frac{(a+b)+c+abc}{1+ab+ac+bc}\\ &=\frac{a(1+bc)+(b+c)}{(1+bc)+(b+c)a}\\ &=\frac{a+\left(\frac{b+c}{1+bc}\right)}{1+a\left(\frac{b+c}{1+bc}\right)}\\ &=a\oplus\left(\frac{b+c}{1+bc}\right)\\ &=a\oplus(b\oplus c). \end{align}$$
But there's more: since $a,b\in (-1,1)$, we have
$$-1<a\oplus b=\frac{a+b}{1+ab}<1$$
by looking at extreme cases, so $a\oplus b\in(-1,1)$.
Yes, it is. And in the expression$$\frac{\frac{a+b}{1+ab}+c}{1+\frac{a+b}{1+ab}c},$$you multiply both the numerator and the denominator by $1+ab$, then you will get$$\frac{a+b+c+abc}{1+ab+ac+bc}.$$Now, do a similar thing to the RHS.