About the notation: Denotes $(a,b)$ the greatest common divisor between $a$ and $b$.
Now, about the exercise, what I did was the following:
Since $(m, n) = (s, n) = 1$,
$ma + nb = sc + nd = 1$ then,
$ma + nb-sc-nd = 1$
$ma-sc + (b-d) n = 1$
let $b-d = j$ then the previous equality I can write it like this: $ma-sc + jn = 1$
I want to get to that $r (ms) + jn = 1$ but I can not think of how to continue to do it.
Besides, is it okay what I did until now?
First lets assume $(m,n)=(s,n)=1$ then there are $x_1,x_2,y_1,y_2\in\mathbb{Z}$ such that $mx_1+ny_1=1$ and $sx_2+ny_2=1$ multiplying them we get $msx+ny=1$ for some integers $x,y$.
Now the other way i.e. lets assume $(ms,n)=1$ and let $(m,n)=k$ then $k|m\Rightarrow k|ms$ but $k|n$ thus $k|(ms,n)$ so $k=1$. Similarly we can show that $(s,n)=1$.