If $H$ is a finite group of order $r$ and $x\in H$, we can see that $x^r=1$. This is because we have the set $\langle x\rangle:=\{1,x,x^2,\ldots,x^r\}\subset H$, and if $x^r\neq 1$ then $\langle x\rangle$ has more elements than $H$, a contradiction. We say that $\min\{t\in\mathbb{N}:x^t=1\}$ is the order of $x$.
If $t$ is the order of $x$, then $r=st+b$ for some $s$ and $0\leq b<t$. We have that $1=x^r=(x^{t})^sx^b=1^sx^b=x^b=1$, and because of the minimality of $t$, we must have that $b=0$. This shows that $t\mid r$. In fact the same proof shows that if $k$ is any number such that $x^k=1$, then $t\mid k$.
Let's look at the problem now. We can see that $AB=\{ab:a\in A,b\in B\}$ has order less than or equal to $mn$ (one reason is that the function $A\times B\to AB$ that sends $(a,b)\mapsto ab$ is surjective and $|A\times B|=mn$).
When exactly is $ab=a'b'$? This happens if and only if $a(a')^{-1}=b(b)^{-1}$ (because $G$ is abelian). We see that $(a(a')^{-1})$ is in $A$, and so $1=(a(a')^{-1})^{m}=(b(b')^{-1})$. Similarly, we have that $1=(b(b')^{-1})^n=(a(a')^{-1})^m$. Because of what we said earlier we have that the order of $b(b')^{-1}$ divides $m$ and $n$ (and the same with the order of $a(a')^{-1}$). Since $m$ and $n$ are relatively prime, these orders must be equal to 1, and so $b=b'$ and $a=a'$. Thus every element of $AB$ is uniquely written as $ab$ with $a\in A$ and $b\in B$, and so we see that the function $A\times B\to AB$ we described earlier is a bijection, and so we're done.
Of course in this proof I proved Lagrange's theorem....
Another proof would consist of showing that the function $A\times B\to AB$ is an isomorphism (although I'm sure you would use Lagrange too).
