How to prove closure property for $x\cdot y = \frac{x + y}{1 + \frac{xy}{c^2}}$?

We need to prove that $$-c\leq\frac{(x+y)c^2}{c^2+xy}<c,$$ which is $$(c-x)(c-y)>0$$ for the right inequality and $$(c+x)(c+y)>0$$ for the left inequality.

Another approach is to write $x=c\tanh a,\,y=c\tanh b$ with $a,\,b\in\Bbb R$ (these are called rapidities), then show $x\cdot y=c\tanh (a+b)$. This proves $G$ is isomorphic to $(\Bbb R,\,+)$.