We can rewrite $u_n(x)=0$ as
$$f_n(x):=\frac{\prod_{k=0}^{n-1}(x+\frac{2k+1}{2})}{\prod_{k=0}^{n-1}(x+k)}=A(n):=\frac{{2n-1 \choose n-1}}{2^{2n-1}}.$$
The graph of $f_n(x)=1+\sum_{k=0}^{n-1} \frac{A_k}{x+k}, A_k>0$ consists of $n+1$ monotonically decreasing ($f_n'(x)<0$) branches with $n-1$ vertical branches over the intervals
$I_j=[-(j+1),-j]$ for $j=0,...n-2$ approaching the end points vertically. In addition, there are two outer branches over $I_{n-1}=(-\infty,-(n-1))$ and $I_{-1}=(0,\infty)$ approaching the horizontal line $y=1$ at $ \pm \infty$ from below for the leftmost branch and from above for the right most branch. Since $A(n)<1$, $f_n(x)=A(n)$ has exactly $n$ negative roots over all the branches $I_j,j=0,..n-1$ except the right most branch and the unique root in $I_{n-1}$ is $-n$. Since $A(n)$ decreases and $f_{n+1}(x) \ge f_n(x)$, the roots in the inner interval increases as we pass from $n$ to $n+1$. The root in the new inner interval $I_{-n}$ is also greater than $-n$. So roots interlace.
(Added) If we let $0 >\lambda_{n,1}>...>\lambda_{n,n}$ be the roots of $u_n(x)$ then interlacing implies for every fixed $k \ge 1$ the $k$th largest root $\lambda_{n,k}$ is increasing so must converge to some $\lambda_k \le -(k-1)$. In this (rare) case, the proof shows that $\lambda_k=-(k-1/2)$ since $A(n) \rightarrow 0$ for large $n$. We also have the $k$th smallest root $\lambda_{n,n-k+1}$ decreasing but they all approach $- \infty$.
It should also be the case that $u_n(x)$ converge locally uniformly on all compact subsets to an entire function with only real zeros at $-(k-1/2),k=1,2,3...$ in the Laguerre-Polya class $\mathcal{L-P}$ but it is essentially the reciprocal of the Gamma function.