Note that $nA+(nm)A = n(A+mA)$. To verify this, if $x\in nA+(nm)A$, then there exists $a,a'\in A$ such that $x = na + nma' = n(a+ma')$; and $a+ma'\in A+mA$. Conversely, $n(a+ma') = na+nma'\in nA+(nm)A$, if $a,a'\in A$.
So we are asking whether $\frac{|n(A+mA)|}{|nA|} = \frac{|A+mA|}{|A|}.$
If $\langle A\rangle$ has no $n$-torsion, then the result follows since multiplication by $n$ is one-to-one (if $na = na'$ then $n(a-a')=0$, so $a-a'\in\langle A\rangle$ is $n$-torsion, so $a=a'$; the same holds for $A+mA\subseteq \langle A\rangle$).
However, if $A$ has $n$-torsion, then you are in trouble. Consider for example $n=5$, $m=3$, $G=\mathbb{Z}/10\mathbb{Z}$, $A=\{1,2\}$. Then $A+3A = \{1,2\}+\{3,6\} = \{4,5,7,8\}$, so the right hand side is $2$. On the other hand, $5(A+3A) = 5\{4,5,7,8\} = \{0,5\}$ and $5A = \{0,5\}$, so the left hand side is $1$.