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.

The statement of the associativity condition is wrong; it should be
$$(a \ast b) \ast c = a \ast (b \ast c) .$$
Expanding the l.h.s. gives
\begin{align}(a \ast b) \ast c &= (a + b - ab) \ast c \\ &= (a + b - ab) + c - (a + b - ab) c \\ &= a + b + c - bc - ca - ab + abc .\end{align}
Now, do the same thing for the r.h.s. and verify that the expressions agree.

We cannot simply assume the existence of an identity and inverse. There are plenty of associative binary operations which lack one or both!
We can pick out the identity element using the definition: We must have, for example, that
$$a \ast e = a$$ for all $a$, and expanding gives
$$a + e - ae = a.$$
Can you find which $e$ makes this true for all $a$?
Likewise, to show that the operation admits inverses, it's enough to produce a formula for the inverse $a^{-1}$ of an arbitrary element $a$, that is, an element that satisfies $a \ast a^{-1} = e = a^{-1} \ast a$. As with the identity, use the definition of $\ast$ and solve for $a^{-1}$ in terms of $a$.

Remark If we glance at the triple product, which by dint of associativity we may as well write $a \ast b \ast c$, we can see the occurrence of the elementary symmetric polynomials in $a, b, c$, which motivates writing that product as $(a - 1) (b - 1) (c - 1) + 1$. Then, glancing back we can see that we can similarly write $a \ast b = (a - 1) (b - 1) + 1,$ which is just the conjugation of the usual multiplication (on $\Bbb R - \{0\}$) by the shift $s : x \mapsto x + 1$, that is, $a \ast b := s(s^{-1}(a) s^{-1}(b))$. Since multiplication defines a group structure on $\Bbb R - \{0\}$, that $\ast$ defines a group structure follows by unwinding definitions. For example, to show associativity, we have
$$(a \ast b) \ast c = s(s^{-1}(s(s^{-1}(a) s^{-1}(b))) s^{-1}(c)) = s(s^{-1}(a) s^{-1}(b) s^{-1}(c)) = s(s^{-1}(a)s^{-1}(s(s^{-1}(b) s^{-1}(c))) = a \ast (b \ast c) .$$
(Note that in writing the third expression without parentheses we have implicitly used the associativity of the usual multiplication.) You can just as well use this characterization to determine the group identity $e$ and the formula for the inverse of an element under $\ast$.
Indeed, this together with analogous arguments for the other group axioms shows that conjugating any group operation by a set bijection defines an operation on the other set.
