Need to prove that $(S,\cdot)$ defined by the binary operation $a\cdot b = a+b+ab$ is an abelian group on $S = \Bbb R \setminus \{-1\}$. 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.

I believe you meant to write $S=\mathbb{R}\backslash\{-1\}$
$0$ is indeed the identity element since for any $a\in S$, $a * 0=a+0+a.0=a$
For $b$ to be the inverse of $a$, we require $a * b=0$.
Hence $a+b+a.b=0$
$b+a.b=-a$
$b(1+a)=-a$
$b=\frac{-a}{1+a}$
which is fine, since $a$ can't be $-1$ (since it's not an element of $S$).
