I'm going to write $(m,n)$ for $\text{gcd}(m,n)$ throughout what follows.
You want to show that $(a,c)(b,c) \mid ab, c$. From the definition of $(m,n)$, it's easy to show that $(a,c)(b,c) \mid ab$ (since $(a,c) \mid a$ and $(b,c) \mid b$). So it really boils down to showing that $(a,c)(b,c) \mid c$.
Since $(a,c), (b,c) \mid c$, you can write $c = r(a,c) = s(b,c)$ for some integers $r$ and $s$. Therefore, to show that $(a,c)(b,c) \mid c$, it's enough to show that $(b,c) \mid r = c/(a,c)$. This is equivalent to showing that $p \nmid (a,c)$ for any prime number dividing $(b,c)$. This follows from $a$ and $b$ being coprime. (Why?)
Take a common divisor $k$ of $a$ and $b$. That means that we can write $a=ka'$, $b=kb'$. Then, if $d=ax+by$,
$$d=a'kx+b'ky=k(a'x+b'y)$$
so we see that $k$ also divides $d$. Since $\gcd(a,b)$ is a common divisor of $a$ and $b$ (namely, the greatest), it also holds the property. Actually, the idea behind this is very simple: the sum of two multiples of the same number is also a multiple of the same number.
Note that $\gcd(a,b)$ is, hence, the least positive integer that can be written as $ax+by$, because every number of the form $ax+by$ must be a multiple of any common divisor of $a$ and $b$, and in particular, must be a multiple of $\gcd(a,b)$ so $ax+by$ can't be smaller than it (being still positive, of course).
Example: let $a=60$, $b=75$. The gcd is $15$, so $ax+by=60x+75y$ must be a multiple of $15$, because it is the sum of two multiples of $15$.
I hope this helps a bit.
Best Answer
By $\rm\color{#c00}H$ere $\color{#c00}{(m,n)\!=\!1}\Rightarrow (ma\!+\!nb,\color{#c00}{mn}) \overset{{\rm\color{#c00}H}}= (ma\!+\!nb,\color{#c00}m)(ma\!+\!nb,\color{#c00}n) = (b,m)(a,n)\ $ by
$$\begin{align}(\color{#0a0}ma+nb,\color{#0a0}m) &\overset{\rm\color{#f6f} R}= (nb,m) \overset{\rm\color{darkorange}L}= (b,m)\\[.2em] (ma+\color{#0af}nb,\ \color{#0af}n) &\overset{\rm\color{#f6f} R}= (ma,n) \overset{\rm\color{darkorange}L}= (a,n)\end{align}\qquad$$
where above we applied gcd $\rm\color{#0a0}{mod}\ \color{#0af}{reduction}$ = $\rm\color{#f6f} R,\ $ and $\ \rm\color{darkorange}L = $ Euclid's Lemma, by $\, \color{#c00}{(m,n)=1}$.