Prove that $a * b = a + b - ab$ defines a group operation on $\Bbb R \setminus \{1\}$ So, basically I'm taking an intro into proofs class, and we're given homework to prove something in abstract algebra. Being one that hasn't yet taken an abstract algebra course I really don't know if what I'm doing is correct here.

Prove: The set $\mathbb{R} \backslash \left\{ 1 \right\}$ is a group under the operation $*$, where:
  $$a * b = a + b - ab, \quad \forall  \,\, a,b \in \mathbb{R} \backslash \left\{ 1 \right\} .$$

My proof structure:
After reading about abstract algebra for a while, it looks like what I need to show is that if this set is a group, it has to satisfy associativity with the operation, and the existence of an identity and inverse element.
So what I did was that I assumed that there exists an element in the set $\mathbb{R} \backslash \left\{ 1 \right\}$ such that it satisfies the identity property for the set and another element that satisfies the inverse property for all the elements in the set. However I'm having trouble trying to show that the operation is indeed associative through algebra since
$$\begin{align}
a(b * c) & = a(b+c) - abc \\
& \ne (a+b)c - abc = (a*b)c \\
\end{align}$$
So in short, I want to ask if it's correct to assume that an element for the set exists that would satisfy the identity and inverse property for the group. Also, is this even a group at all since the operation doesn't seem to satisfy the associativity requirements.

Associativity is satisfied:
$$a*(b*c)=\\a*(b+c-bc)=\\a+b+c-bc-a(b+c-bc)=\\a+b+c-ab-bc-ac+abc$$
and $$(a*b)*c=\\(a+b-ab)*c=\\a+b-ab+c-(a+b-ab)c=\\a+b+c-ab-bc-ac+abc$$
