So basically this proof centers around proving that (S,*) is a group, as it's quite easy to see that it's abelian as both addition and multiplication are commutative. My issue is finding an identity element, other than 0. Because if 0 is the identity element, then this group won't have inverses.
The set explicitly excludes -1, which I found to be its identity element, which makes going about proving that this is a group mighty difficult.
Best Answer
I assume you mean $S=\mathbb R\setminus \{-1\}$.
Take $f:S\to \mathbb R^*$ given by $f(x)=x+1$. This is a bijection.
Now note that for $a,b\in S$, we have $a*b= a+b+ab=(a+1)(b+1)-1=f^{-1}(f(a)f(b))$. What $f$ does is to rename the elements of $S$ as elements of $\mathbb R^*$ and operate there. So $S$ is a group because it has been forced to be isomorphic to $\mathbb R^*$ via $f$. That's all there is to it.
For instance, $0\in S$ is the identity because $f(0)=1$ and $1$ is the identity of $\mathbb R^*$.
This is an example of a pullback. See https://math.stackexchange.com/a/373743/589