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 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

In general for any associative ring $R$, the circle operation $x \circ y=x+y+xy$ defines a monoid structure on the underlying set of $R$, and the invertible elements in this monoid form a group called the adjoint group of the ring $R$, let us denote it by $Q(R)$.

If $R$ has an identity for the multiplication, $Q(R)$ is isomorphic to the group of units of $R$ under the mapping $x \mapsto 1+x$, equivalently the set $\{\,y-1 \in R \mid y \mbox{ is a unit in }R\,\}$ forms a group under our circle operation.

If we are working with a field $\mathbb R$, we have only to exclude $0-1=-1$ from this set. In particular, the elements of $\mathbb{R}\setminus\{-1\}$ form a group under our circle operation.

I think the identity has to be 0: $a*e = a+e+ae=a \Rightarrow e+ae=0 \Rightarrow e(1+a)=0 \Rightarrow e=0 $. The inverse of $a$ is: $a*a^{-1}=0 \Rightarrow a+a^{-1}+aa^{-1}=0 \Rightarrow a^{-1} = -\frac{a}{1+a}$. Since $a \neq -1$, so for all $a$, $a^{-1}$ always exists.