$\textbf{Problem setup}$:
Let $F : R_{\tau} \rightarrow R_{\tau}$ be a biholomorphism onto its image (i.e. $F : R_{\tau} \rightarrow V$ is a holomorphic function with holomorphic inverse, here $V = F(R_{\tau}) \subset R_{\tau}$ is open), and $R_{x} = \{ w \in \mathbb{C} : \Re(w) > x \}$ for any $x \in \mathbb{R}$. Here $\tau > 0$ is some fixed real number.
Moreover assume that $F$ extends holomorphically to a slightly larger right half plane $R_{\tau’}$ with $0<\tau’ < \tau$ and is again a biholomorphism onto its image.
Suppose $F(w) = w + 1 + \eta(w^{-\frac{1}{n}})$ (holds in $R_{\tau’}$), where $\eta(\zeta) = b_1 \zeta + b_2 \zeta^2 + ...$, is some analytic function in a neighbourhood of $0$, defined on $|\zeta| < r$ for some fixed $r > 0$. Also assume $|\eta(\zeta)| \leq \frac{1}{2}$ for all $|\zeta| < r$.
Now suppose $\tau > \frac{1}{r^n} + 1$ is fixed. For any $x \geq \tau$ real define $S_{x} = \{ x \leq \Re(w) \leq x+1 \}$. Also set $E_{x} = \{ \Re(w) = x \}$.
For any $w_0 \in R_{\tau} \cup E_{\tau}$, set $w_m = F^m(w_0)$. Then due to the form of $F$, we necessarily have that $w_{m} - w_{m-1} \rightarrow 1$, and the convergence (in $m$) is uniform for $w_0 \in R_{\tau} \cup E_{\tau}$.
Now let $x_0 = \tau$, $\zeta_{m} = F^m(x_0)$ and $x_m = \Re(\zeta_{m})$. Define $\Omega_{m} = F^{-m}(R_{x_m})$.
$\textbf{Problem statement}$:
It is apparently a fact that (at least for sufficiently large $m$), $\Omega_{m} = \bigcup_{i=-\infty}^m F^{-i}(S_{x_{m}})$.
It is clear to me that $\bigcup_{i=-\infty}^m F^{-i}(S_{x_{m}}) \subset \Omega_{m}$, but I do not see any reason to believe the other inclusion should hold.
It is likely I am missing a step involving using that $w_{m} - w_{m-1} \rightarrow 1$ uniformly.