Suppose, to the contrary, that there is an isomorphism $\phi:(\mathbb Q,+)\to(\mathbb Q_{>0},\times)$. Then, there must be an $a\in\mathbb Q$ such that $\phi(a)=2$. Thus, $$\phi(a)=\phi(a/2+a/2)=\phi(a/2)\times \phi(a/2)=2 \, ,$$ which is absurd, since $2$ does not have a square root in $(\mathbb Q_{>0},\times)$.

More conceptually, $(\mathbb Q,+)$ is a divisible group, whereas $(\mathbb Q_{>0},\times)$ is not. A divisible group is a group (usually assumed to be commutative) where for every positive integer $n$ and $g\in G$, there is a $h\in G$ such that $h^n=g$. In additive notation, this condition would be written $nh=g$. The above proof exploits the fact that $(\mathbb Q_{>0},\times)$ is not divisible; specificially, if $n=2$ and $g=2$, then there is no $h\in\mathbb Q^+$ such that $h^n=g$.

Suppose, to the contrary, that there is an isomorphism $\phi:(\mathbb Q,+)\to(\mathbb Q_{>0},\times)$. Then, there must be an $a\in\mathbb Q$ such that $\phi(a)=2$. Thus, $$\phi(a)=\phi(a/2+a/2)=\phi(a/2)\times \phi(a/2)=2 \, ,$$ which is absurd, since $2$ does not have a square root in $(\mathbb Q_{>0},\times)$.

More conceptually, $(\mathbb Q,+)$ is a divisible group, whereas $(\mathbb Q_{>0},\times)$ is not. A divisible group is a group (usually assumed to be commutative) where for every positive integer $n$ and $g\in G$, there is a $h\in G$ such that $h^n=g$. In additive notation, this condition would be written $nh=g$. The above proof exploits the fact that $(\mathbb Q_{>0},\times)$ is not divisible; specificially, if $n=2$ and $g=2$, then there is no $h\in\mathbb Q_{>0}$ such that $h^n=g$.