The cyclic group $G$ is isomorphic to the additive group of $\Bbb Z/rs\Bbb Z$, so this is just the Chinese remainder theorem for the coprime moduli $r$ and $s$ (in the statement for rings $\Bbb Z/n\Bbb Z$, but only considering their additive structure).
Concretely, the elements of $G$ of order dividing $r$ are generated by $x^s$ and vice versa, and among the $r$ elements of order dividing $r$ there is one, say $y$, such that $z=y^{-1}x$ has order dividing $s$. One has $x=yz$ and $y,z$ commute (they are both in the group $G$ generated by $x$); if either the order of $y$ were a strict divisor of $r$ or the order of $z$ were a strict divisor of $s$ then it would follows that the order of $x$ is a strict divisor of $rs$, which is false, so the orders of $y,z$ are respectively exactly $r,s$. Concretely you can find $y,z$ by writing $1=\gcd(r,s)=ar+bs$ using the extended Euclidean algorithm; then $y=x^{bs}$ and $z=x^{ar}$.