First verify * is a binary operatin.

If a and b $\in$ G then $a*b\bigg(=\dfrac{ab}{2}\bigg)$ is a nonzero real number,

$\therefore a*b \in G$

Associativity : Since $(a*b)*c=\bigg(\dfrac{ab}{2}\bigg)*c=\dfrac{(ab)c}{4}$

$a*(b*c)=a*\bigg(\dfrac{bc}{2}\bigg)=\dfrac{a(bc)}{4}=\dfrac{(ab)c}{4}$

$\therefore *$ is associative.

The number 2 is the identity in G and $a \in G$ then

$$a*2=\dfrac{(a)(2)}{2}=a=\dfrac{(2)(a)}{2}=2*a$$

Finally, if $a \in G$ then $a'=\dfrac4a$ is an inverse of a.

$\because a*a'=a*\dfrac4a=\dfrac{a(4/a)}{2}=2\dfrac{(4/a)(a)}{2}=\dfrac4a*a=a*a'$

Since $a*b=b*a \forall a, b \in G$ we conclude that G is an abelian group.