The first issue is that there no reason a priori that such a domain $U$ exists. Consider for example the Cauchy distribution:
$$
f(x) = \frac1{\pi(1+x^2)}
$$
At best, you can define $C$ on the imaginary line, for $\omega\in\mathbb R$:
$$
C(i\omega) = -|\omega|
$$
There is therefore no analyticity to consider.
In general you can start from the moment generating function:
$$
M(t) = \int f(x)e^{tx}dx
$$
You know that there is $t_-\leq0\leq t_+$ such that in the domain $U =\{t\in\mathbb C|t_-<\Re t<t_+\}$, $M$ is defined and analytic. It is the continuous analogue of the inner and outer radii of convergence for Laurent series. The previous example shows that you can have $t_+=t_-=0$. Now assume in the following that $t_-<t_+$. Since $M$ is defined and analytic on $U$, $C = \ln M$ is defined and analytic on $U$ minus the isolated zeros of $M$ ($M\neq 0$) which is still an open set. In fact, by positivity of $f$, you know that $C$ is defined on the interval $(t_-,t_+)$ where it is convex. Furthermore, since $M$ is continuous and $C$ is non zero on the interval, $M$ is still non zero in a neighborhood of the interval, so $C$ is defined on such a neighbourhood (i.e. zeros of $M$ are at a finite distance from $(t_-,t_+)$). You can even extend the domain of definition if you want to by analytic continuation.
For example, consider the Laplace distribution:
$$
f(x) = \frac12e^{-|x|}
$$
with $H$ the Heaviside function. You have:
$$
M(t) = \frac1{1-t^2}
$$
with $t_\pm = \pm1$ and:
$$
C(t) = -\ln(1-t^2)
$$
You can extend both $M,C$ to the entire complex plane minus the poles at $\pm t$.
In general, if you want to reduce to the Laplace transform, you need to assume that the random variable is non negative almost surely, i.e. $f$ is supported on $\mathbb R_+$. You therefore automatically have $t_- = -\infty$ and you are guaranteed to have a non trivial domain where $C$ is analytic.
