(The set of all non-zero rationals is often denoted, instead, by $\mathbb{Q}^*$, but I'll stick with your notation.)
Abelian ?:
Let $a, b \in \mathbb{Q}^+$
Then $\;\;a*b = \dfrac {ab}{2} = \dfrac {ba}{2} = b * a$
$\quad\quad\quad$(...since "normal" multiplication of two rational numbers is commutative: $a\cdot b = b\cdot a$.)
Hence, the group is abelian.
The order of this group, having showed it is a group, is the order of $\mathbb{Q}^+ = |\mathbb{Q}^+| =|\mathbb{Z}|$, both of which you know are infinite.
If the order were some finite $n \in \mathbb{N}$ (which we'll simply call "$n$"), then for every element $g \in \mathbb{Q}^+$, then it would have to be true that $g^n = e$, where $e$ is the identity for $*$ in $\mathbb{Q}^+$.
Clearly, no such number $n$ exists so that $g^n = e$ is true for all elements in $\mathbb{Q}^+$. Hence, the order of your group cannot be finite, and therefore must be infinite.