Note that $u_n$ is the number of ordered pairs $\langle k,\ell\rangle$ of non-negative integers such that $k+\ell=n$ and $k\le\ell$. It’s not hard to show that
$$u_n=\left\lfloor\frac{n}2\right\rfloor+1$$
for all $n\in\Bbb N$. Thus,
$$\begin{align*}
U(x)&=\sum_{n\ge 0}\left(\left\lfloor\frac{n}2\right\rfloor+1\right)x^n\\
&=\sum_{n\ge 0}\left\lfloor\frac{n}2\right\rfloor x^n+\sum_{n\ge 0}x^n\\
&=\sum_{n\ge 0}nx^{2n}+\sum_{n\ge 0}nx^{2n+1}+\sum_{n\ge 0}x^n\\
&=(1+x)\sum_{n\ge 0}nx^{2n}+\sum_{n\ge 0}x^n\\
&=(1+x)\sum_{n\ge 1}nx^{2n}+\sum_{n\ge 0}x^n\\
&=(1+x)x^2\sum_{n\ge 1}n\left(x^2\right)^{n-1}+\sum_{n\ge 0}x^n\\
&=\frac12(1+x)x\sum_{n\ge 1}n\left(x^2\right)^{n-1}(2x)+\sum_{n\ge 0}x^n\;.\tag{1}
\end{align*}$$
Now find closed forms for the two summations in $(1)$, and take the reciprocal of the resulting expression. You should know a closed form for the second summation, and I’ve manipulated the first into a form that makes it quite easy to find a closed form for it; just use a little calculus.