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.
 A: 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
A: 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. 
A: 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. 
A: 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 $S$).
