For real numbers $x > 1$, which forms the set $G$, it is given that the operation on $a,b$, being $a\ast b$, results in $ab - a - b + 2$ (where $ab$ is the ordinary multiplication of $a$ and $b$).
Given that $(G,\ast)$ is known to be an Abelian Group, how can it be shown that the group is closed, and has an identity element?
On
If it is already given to be a group then what's left to prove. So I am assuming that you were supposed to check if It is a group?
You need to establish that $a*b>1$. observe that $a*b=(a-1)(b-1)+1$. Since both $a$ and $b$ are greater than $1$ we get the result.
Let $e$ be the identity element. Therefore for all $a\in G$ we should have. $a*e=e*a=a$.
This means $ae-e-2a+2=0$. This suggests that $e=2$ can be the identity element. Now you may verify it.
$G$ is closed if $a$, and $b$ $\in G$ implies: $a*b \in G$. $a*b = ab - a - b + 2 > 1 \iff ab - a - b + 1 > 0 \iff (a - 1)(b - 1) > 0$ which is true since $a > 1$ and $b > 1$.
$e$ is an identity of $G$ if: $x*e = x$ for all $x \in G \iff xe - x - e + 2 = x \iff (e - 2)(x - 1) = 0$ for all $x > 1$. This implies: $e = 2$.