This is just a partial answer, but anyway:
( I use $s[n,k]$ for $s^n_k$ since I just copy-pasted from Mathematica:
I do not know if this helps, or if you already know, but you have the recurrence
$$s[n- 1,k + 1] = s'[n, k] - x \cdot s'[n - 1, k]$$
Note, by induction, we assume $s[n,k]$ and $s[n-1,k]$ to have interlacing roots.
By thm 1.47 in http://arxiv.org/abs/math/0612833 (look into this work),
we know that the derivatives have interlacing roots as well.
Now, the recursion is quite similar to many other that appear in the link above,
so it might not be too hard to prove stability.
I show/sketch below that the operator $f \mapsto A f - x f'$
always produce roots interlaced (and to the right) of the roots of $f$,
provided $A>\deg f$ and that all roots of $f$ are positive.
(We may assume leading term of $f$ has positive coefficient).
Clearly if f has a root of multiplicity m, then the result will have the same root with multiplicity m-1, so the problem is essentially to ensure that no new multiple roots may appear.
Assume now we have two consecutive roots $0\le a \le b$ of $f$ and that $f$ is positive between these.
The derivative of $f$ this first positive, and then negative in $[a,b]$,
thus $-x f'$ is first negative and then positive.
Hence, $A f - x f'$ is negative in $a$, and positive in $b.$
Thus, we have a root of $A f - x f'$ in the interval $[a,b]$.
(Mutatis mutandis for the case when f is negative between $a$ and $b$).
Now, if the degree of $f$ is even, $f$ is positive to the right of its largest root.
The derivative is also positive here (since even degree poly), so $x f'$ is positive.
Hence, $A f - x f'$ is negative in this point. However, notice this is a polynomial with even degree, and with a leading coefficient! Hence, it must eventually cross the real line and grow to infinity.
(Mutatis mutandis for the case when f is odd).
This proves that $f \mapsto A f - x f'$
produces interlacing roots, and eventual multiplicities are decreased, no new multiple roots can be introduced.