Can the modulus of an entire function of the form $ze^{g(z)}$ tend to $1$ in the upper half-plane? Does there exist an entire function $g$ such that $|ze^{g(z)}| = 1 + o(1)$ as $\text{Im}(z)\to+\infty$?
Question can be reformulated in terms of $u(z) = \text{Re}(g(z))$. Namely, does there exist a harmonic in $\mathbb{C}$ function $u$ such that $u(z) = -\log(|z|) + o(1)$ as $ \text{Im}(z)\to+\infty $? In this case the function $-u$ is positive in some half-plane $\{\text{Im}(z) > N\}$ and therefore $-u$ admits the Herglotz representation without the linear term there.

Yes. For $\Im(z) > 0$ let
$$g(z) = -\log z + \int_{i\infty}^z \frac{e^{is}}{s}ds$$
$\int_{i\infty}^{x+iy} \frac{e^{is}}{s}ds= \int_{i\infty}^{iy}\frac{e^{is}}{s}ds
-i \frac{e^{is}}{s}|_{iy}^{x+iy}-i\int_{iy}^{x+iy}\frac{e^{is}}{s^2}ds$ converges to $0$ uniformly as $y\to \infty$.
So $z e^{g(z)} = e^{o(1)}$ as $\Im(z)\to \infty$.
$g(z)$ extends to an entire function because $$g'(z)= \frac{e^{iz}-1}z$$
