You need to show that for any two elements $a, b$ in $G$, $a*b = b*a$.
$$a*b = \dfrac{\color{blue}{a\cdot b}}2 = \dfrac{\color{blue}{b\cdot a}}2 = ba$$
Because "normal" multiplication on the real numbers is commutative, we know $\color{blue}{a\cdot b = b\cdot a}$. Hence, we have commutativity.
Now you need to establish that $G$ is in fact, a group:
Associative? Check whether, given any $3$ elements $a, b, c$ in $G$, is it true that $$a*(b*c) = (a*b)*c\;\;?$$
Existence of an Identity element $e$ such that for all $a \in G$, $a*e = e*a = a$? Let's check out whether $2$ meets this criteria: $$a*2 = \dfrac {a \cdot 2}2 =\color{blue} a = \dfrac{2\cdot a}2 = 2*a$$
Closure under inverses?: For every element $a\in G$, does there exist an element $a^{-1}$ such that $aa^{-1} = a^{-1}a = 2= e$? Let's check out whether $\frac 4a = a^{-1}$: $$a*\frac 4a= \frac {a\cdot \frac 4a} 2 = \frac 42 = \color{blue}2 = \frac{\frac 4a \cdot a}2 = \frac 4a*a$$
Once you confirm that associativity holds, we'll see that $[G, *]$ is a commutative group, i.e. and abelian group.
Best Answer
At the request of the OP, I will show a direct proof. Let $G$ be the set of positive rational numbers. $G$ is a group under mutiplications. Let $\Pi$ be the set of prime numbers. We claim $G$ is a free abelian group with a basis $\Pi$.
Let $r \in G$. there exist positive integers $a, b$ such that $r = \frac{a}{b}$. Since $a$ and $b$ can be written as products of prime numbers, $G$ is generated by $\Pi$.
Let $p_1,\dots,p_r$ be distict prime numbers. Suppose $p_1^{n_1}\cdots p_r^{n_r} = 1$, where all $n_i$ are integers. It suffices to prove that all $n_i = 0$. If all $n_i \ge 0$, clearly all $n_i = 0$. Hence we assume not all $n_i \ge 0$. Without loss of generality, we can assume that $n_1, \dots, n_k \ge 0$ and $n_{k+1},\dots n_r < 0$. Then $p_1^{n_1}\cdots p_k^{n_k} = p_{k+1}^{-n_{k+1}}\cdots p_r^{-n_r}$. But this is a contradiction because $\mathbb{Z}$ is a UFD.