The same proof works if one take $A$ to be non-normal instead of being normal.
One has $AB/B \cong A/A \cap B$ is soluble since it is a homomorphic image of a soluble one. As $B$ is soluble and the class of soluble groups is closed under taking extensions of its members, $AB$ is soluble.
If I remember, $AB$ is also soluble, if one drop the normality condition and assume that both $A$ and $B$ are nilpotent.
The "standard" proof:
Consider, for each $i$, the subgroups $G_iN/N$ of $G/N$. It is straight-forward to show that $G_iN$ is normal in $G_{i+1}N$ (and thus $G_iN/N$ is normal in $G_{i+1}N/N$).
Now $\dfrac{G_{i+1}N/N}{G_iN/N} \cong G_{i+1}N/G_iN$.
Taking any $x,y \in G_{i+1}$, and $n,n' \in N$, we see that:
$[xn(G_iN),yn'(G_iN)] = [xn,yn']G_iN$, and since $N \lhd G$, $[xn,yn'] \in [x,y]N$, so that:
$[xn,yn']G_iN = [xn,yn']NG_i = ([xn,yn']N)G_i = $
$([x,y]N)G_i = [x,y]NG_i = [x,y]G_iN = ([x,y]G_i)N$.
Since $G_{i+1}/G_i$ is abelian, $[x,y]G_i = G_i$ so $[xn,yn']G_iN = G_iN$,so that
$G_{i+1}N/G_iN$ is abelian, as desired.
(here, $[x,y] = xyx^{-1}y^{-1}$ and we are using the fact that for a group $G$, $G$ is abelian iff $[x,y] = e$ for all $x,y \in G$).
It should be noted that some of the quotients $G_{i+1}N/G_iN$ may be trivial, resulting in a shorter subnormal series for $G/N$.
Best Answer
Hint: construct an injective homomorphism from $G/(K \cap N)$ to $G/K \times G/N$ and use that products and subgroups of solvable groups are again solvable.