Proving if the set is a group under the following operation So I’ve started learning about group theory this semester (actually just this week) and I’m a total newbie in the field. I’ve got the following set and I need to prove if it’s a group with respect to the operation stated below: 
$$  G := \mathbb R \setminus \{1\} $$
under the operation: 
$$a \circ b =  a +b-ab$$
for $$ a,b \in G $$
So far I've proved that It's associative and the identity is the number $0$, but I can't manage to prove the closure property and also can't find the inverse.
Any help in any form would be highly appreciated.

I think an easy way to prove the closure property is to see that, $a\circ b \ne 1, \forall a, b \in \mathbb R \setminus \{1\}$. 
Using the identity $a+b-ab-1=(1-b)(a-1)$, we can see that $a+b-ab \ne 1 \Leftrightarrow  (1-b)(a-1) \ne 0, \forall a, b \in \mathbb R \setminus \{1\}$

The inverse of $a$ goes for resolving $a\circ a^{-1} = 0$, because $0$ is th identity element of the group $G$. So, $a\circ a^{-1} = 0 \Leftrightarrow a + a^{-1} - aa^{-1} = 0 \Leftrightarrow a + a^{-1} - aa^{-1} - 1 = -1 \Leftrightarrow (1-a^{-1})(a-1) = -1$.
Since $a, a^{-1} \ne 1$:
$(a^{-1}-1) = \frac{-1}{1-a} \Leftrightarrow a^{-1} = 1 + \frac{1}{a-1} \Leftrightarrow a^{-1} = \frac{a}{a-1}$
