I would be willing to wager vast sums of money that this function has the unit circle as a boundary of essential singularities. As $x\to 1$ it diverges like $u(x) \sim c (1-x)^{-2}\log{(1-x)}$.

Edit: Hi Ben, let me say a bit more. Write $N(t)$ for the number of zeros of the Riemann zeta function in the strip $0 < Re(s) < 1$ with imaginary part between $0$ and $t$. According to Riemann we have $N(t)=\frac{t}{2\pi}\log{(\frac{t}{2\pi e})}+S(t)$ where $S(t)$ satisfies $S=O(\log{t})$. This function $S(t)$ captures the fine variation in the distribution of the zeros of the zeta function. It is quite random: Selberg proved unconditionally, in a triumph of analytic number theory, that $(\log{\log{t}})^{-\frac{1}{2}}S(t)$ has the distribution of a Gaussian random variable with mean zero and variance one!

Now, what if you try to invert this? Write $\gamma_n$ for the imaginary part of the $n$th zero, ordered by height. After inverting the above functions you'll get something like $\gamma_n=f(n)+r(n)$ where $f$ is essentially the valute of $t$ such that of $n=\frac{t}{2\pi}\log{(\frac{t}{2\pi e})}$, and $r$ is some chaotically varying error term. Splitting $u$ as $u_f(x) + u_r(x)$ accordingly, I am sure that $u_f(x)$ has a nice analytic continuation, but for all intents and purposes $u_r(x)$ looks like a power series with "random" coeffecients. Such things generally do not have good analytic continuations

That being said, this is not a proof of my claim, and I would be delighted to see one.