Is there a non-analytic function $g$ such that $\exp(s g(z))$ is always analytic? Does there exist a complex function $g(z)$ that is not analytic on the unit disk, such that $\exp(s g(z))$ is analytic on the unit disk for all complex numbers $s$? What about for some continuous range of $s$ (e.g. all real numbers, or all positive numbers, or all $s \in [0,1]$)?
For example, if we only consider $s=1$, then $g(z) := \log(z)$ is not analytic on the unit disk, but $\exp(\log(z)) = z$ obviously is. But if we allow $s$ to vary, then $\exp(s \log(z)) = z^s$ is only analytic on the unit disk if $s$ is a natural number, so $g(z) = \log(z)$ doesn't work.
 A: First of all, contrary to my original believe, one really needs this additional parameter $s$ for the statement having a chance to be true. In fact $g$ being continuous would be enough.
Example of non-analytic function if we consider only $s=1$:
Without the condition with the parameter $s$ we have the counterexample
$$ g(z) = \begin{cases} 0,& z=0, \\ 2\pi i,& z \neq 0. \end{cases} $$
Proof if we assume in addition to the conditions listed by the OP that $g$ is bounded or continuous:
One can also show the following version

Let $\Omega\subseteq \mathbb{C}$ be an open set, $\varepsilon>0$ and $g: \Omega \rightarrow \mathbb{C}$ bounded or continuous such that $\Omega \rightarrow \mathbb{C}, z \mapsto e^{sg(z)}$ is analytic for all $s\in (0;\varepsilon)$, then $g$ is analytic.

If $g$ is continuous and $g$ does not vanish anywhere, then we can write locally $g(\zeta) = \ln(e^{g(\zeta)})$ in a neighborhood of $z$, where $\ln$ is the branch of the logarithm that maps $e^{g(z)}$ to $g(z)$. As it is the composition of analytic functions, it is analytic. However, analyiticity is a local property, thus, $g$ is analytic.
Now if $g(z) = 0$, then we consider $h(w) :=g(w)+1$. Then we have $e^{h(w)} = e \cdot e^{g(w)}$, which is analytic and $h(z) =1$. Thus, $h$ is analytic in a neighborhood of $z$. Then clearly $g(w) = h(w) -1$ is analytic as well in a neighborhood of $z$. Therefore, if $g$ continuous we get that $g$ is analytic.
Next we want to show that if $g$ is assumed to be bounded, then the conditions of the OP force $g$ to be continuous as well. Let $z\in \mathbb{C}$ and pick any sequence $(z_n)_{n\in \mathbb{N}}$ such that $z_n \rightarrow z$. Then we have (as $e^g$ is continuous)
$$ e^{g(z)} = \lim_{n\rightarrow \infty} e^{g(z_n)}.$$
Taking the absolute value, we get
$$ e^{\text{Re}(g(z))}=\vert e^{g(z)} \vert = \lim_{n\rightarrow \infty} e^{\text{Re}(g(z_n))}. $$
As $\ln:(0;\infty) \rightarrow \mathbb{R}$ is continuous, we get
$$ \text{Re}(g(z)) = \lim_{n\rightarrow \infty} \text{Re}(g(z_n)).$$
Thus, we get
$$ e^{i\text{Im}(g(z))} = \frac{e^{g(z)}}{e^{\text{Re}(g(z))}} =
 \frac{ \lim_{n\rightarrow \infty} e^{g(z_n)}}{\lim_{n\rightarrow \infty} e^{\text{Re}(g(z_n))}} = \lim_{n\rightarrow \infty} \frac{e^{g(z_n)}}{e^{\text{Re}(g(z_n))}} = \lim_{n\rightarrow \infty} e^{i\text{Im}(g(z_n))}.$$
Now the problem is to get rid of the potential differences of $2\pi$. If you assume that $g$ is bounded, then this is quite easy. Then we can pick subsequences, where $i \text{Im}(g(z_{n_k}))$ actually converges. If you have different limits (after removing the angle) $2\pi m$ and $2\pi \ell$, then we can just pick $s$ so small that $s2\pi m$ and $s2\pi \ell$ are both smaller than $2\pi$ and then they need to coincide. This means $g$ is continuous (as both $\text{Re}(g)$ and $\text{Im}(g)$ are continuous) and hence by a previous argument $g$ is also analytic.
Replacing boundedness with more parameters:
In order to conclude that $g$ is analytic without continuity/boundedness of $g$, one needs to handle the case, when $g$ blows up. We can handle this case if we can choose the parameter $s$ in a slightly larger set:

Let $\Omega\subseteq \mathbb{C}$ be an open set, $\varepsilon>0$ and $g: \Omega \rightarrow \mathbb{C}$ such that $\Omega \rightarrow \mathbb{C}, z \mapsto e^{sg(z)}$ is analytic for all $s\in B(0;\varepsilon)=\{ z\in \mathbb{C} \ : \ \vert z \vert <\varepsilon \}$, then $g$ is analytic.

Let us define
$$ G: \Omega \times B(0;\varepsilon) \rightarrow \mathbb{C}, (z,s) \mapsto e^{sg(z)}. $$
First let us note that $G$ is separately holomorphic and hence by Hartogs' theorem $G$ is continuous. Let $z_0\in \Omega$, we just want to show that in a neighborhood of $z_0$ the function $g$ is bounded, then we are done by the argument we gave above.
Let us fix some closed ball $B=\overline{B(z_0, \eta)} \subset \Omega$ (where $\eta>0$). Assume by contradiction that $g$ is not bounded on $B$. Then there exists a sequence $(\zeta_n)_{n\in \mathbb{N}} \subseteq B$ such that $\vert g(\zeta_n) \vert \rightarrow \infty$. As $B$ is compact, we can assume wlog that $\zeta_n \rightarrow \zeta$ for some $\zeta \in B$. However, then we get the contradiction
$$ 1 = e^0 = G(\zeta, 0) = \lim_{n\rightarrow \infty} G(\zeta_n, \frac{1}{\vert g(\zeta_n) \vert})= e^1 =e. $$
