I want an example of holomorphic map $f : \Omega \to \Bbb C$ where $\Omega$ is the half-plane $\mathrm{Re}(s) > 1$, such that there is no sequence $(a_n)$ of complex numbers with $$f(s) = \sum_{n \geq 1} a_n/n^s, \quad\forall s \in \Omega$$
I've read that $(s − 1)\zeta(s)$ is such an example, but I'm not sure how to prove it.
If a function can be represented by a Dirichlet series, $$f(s) = \sum_{n = 1}^{\infty} \frac{a_n}{n^s}$$ for $\operatorname{Re} s > \sigma_0$, then we have $$\lim_{\operatorname{Re} s \to +\infty} f(s) = a_1 \tag{$\ast$}$$ uniformly in $\operatorname{Im} s$. (And the convergence to $a_1$ is exponentially fast.) If the Dirichlet series converges absolutely for $\operatorname{Re} s = \sigma_1$, then for $\sigma = \operatorname{Re} s > \sigma_1$ we can estimate \begin{align} \lvert f(s) - a_1\rvert &= \Biggl\lvert \sum_{n = 2}^{\infty} \frac{a_n}{n^s}\Biggr\rvert \\ &\leqslant \sum_{n = 2}^{\infty} \frac{\lvert a_n\rvert}{n^{\sigma}} \\ &= \sum_{n = 2}^{\infty} \frac{\lvert a_n\rvert}{n^{\sigma_1}}\cdot \frac{1}{n^{\sigma - \sigma_1}} \\ &\leqslant \frac{1}{2^{\sigma - \sigma_1}}\sum_{n = 2}^{\infty} \frac{\lvert a_n\rvert}{n^{\sigma_1}}\,. \end{align}
So a function that does not satisfy $(\ast)$, be it that there are paths (with real part tending to $+\infty$) along which $f(s)$ oscillates, or that $\lvert f(s)\rvert$ is unbounded along such a path, cannot be represented by a Dirichlet series.
Since \begin{gather} \lim_{\operatorname{Re} s \to +\infty} \lvert (s-1)\zeta(s)\rvert = +\infty \quad \text{and}\\ \lim_{\operatorname{Re} s \to +\infty} \lvert e^s\rvert = +\infty \end{gather} we see that neither of those functions can be represented by a Dirichlet series.
The function defined by $f(s) = 1/s$ for $\operatorname{Re} s > 1$ satisfies $(\ast)$ - with $a_1 = 0$ - uniformly in $\operatorname{Im} s$, but since the convergence to $0$ is too slow - it's not exponential - this function cannot be represented by a Dirichlet series either.