Yes, because $u_n(x)$ switches sign between each consecutive pair in
$x=0,-1,-2,-3,\ldots,1-n$, and $u_n(-n) = 0$.

In general, if $P,Q$ are continuous functions each with $n$ simple roots
in an interval $I$, and those roots interlace, then the same argument gives
at least $n-1$ roots of $P-Q$ in $I$, and then an extra root (or even two)
might be forced by the values of $P-Q$ at the endpoints of the interval.
Here $P$ and $Q$ are $2^{n-1} \prod_{k=0}^{n-1} \, (2x+2k+1)$ and
${2n-1 \choose n-1} \prod_{k=0}^{n-1} \, (x+k)$, and $I = (-\infty,0)$.