Let $D$ be a simply connected domain on the complex plain $\mathbb{C}$ with analytic boundary. Let $\Gamma$ be a simple curve that connects two discrete points $a$ and $b$ on the boundary of $D$, so that it splits $D$ into two disjoint parts, say $D_1$ and $D_2$. If we know that $f:\overline{D}\to\overline{\mathbb{C}}$ is holomorphic on $\overline{D}\setminus\{a\}$ with essential singularity at $z=a$ and its modulus $|f|$ is strictly less than $\tau>0$ on $D_1$, constantly equal to $\tau$ on $\Gamma$, does it hold that $|f|$ is strictly bigger than $\tau$ in $D_2$? Or if not in whole $D_2$, at least in a subdomain of it, say $D_3$, that part of its boundary is $\Gamma$?
By the maximum modulus principle, we can find a point $z_0$ in $D_2$ such that $|f(z_0)|>\tau$, since on any disk $D(\zeta,r)$ with center $\zeta\in\Gamma$ and suitable radius $r>0$, $|f(z)|<\tau$ in $D_1\cap \overline{D(\zeta,r)}$, $|f(z)|=\tau$ on $\Gamma\cap D(\zeta,r)$. The function $f$ being holomorphic in $D(\zeta,r)$ cannot attain its maximum in $D(\zeta,r)$, so there is a $z_0$ in $D_2\cap D(\zeta,r)$ such that $|f(z_0)|>\tau$. By continuity, there is a small disk around $z_0$ on which the modulus $f$ is strictly bigger than $\tau$. Now we can repeat the same argument for another $\zeta\in\Gamma$ and for another $r'>0$ to end up with many such small disks as described above, but does this disks fill up $D_2$?
EDIT:
Consider the set $$D_3=\{z\in D:\,|f(z)|>\tau\}.$$ By the maximum modulus principle $D_3\neq\emptyset$, $D_3\cap D_1=\emptyset,$ since $|f(z)|<\tau$ for every $z\in D_1$, $D_3\cap\Gamma=\emptyset,$ since $|f(z)|=\tau$ for every $z\in\Gamma$. Thus $D_3\subset D_2$. By continuity $D_3$ is open, since for every $z\in D_3$ there is a disk around $z$ contained in $D_3$.
So the question is:
Is $D_3=D_2$? I think the answer is yes.
(I take $\tau=1$ in this answer.) At least without the requirement for an essential singularity, no, $\lvert f(z) \rvert > 1$ does not have to hold on $D_2$. However, there is always an open neighborhood $U$ of $\Gamma$ such that this inequality holds on $U \cap D_2$ where $U$ does not depend on $f$. This follows from an generalization of Schwarz’ reflection principle for smooth analytic curves: A part of $U_1 \subseteq D_1$ such that $\Gamma \subset \overline{U}_1$ can be “reflected” in $\Gamma$. If $\sigma$ is this reflection (an anti-holomorphic involution fixing $\Gamma$ pointwise) then $f$ satisfies $\overline{f(\sigma z)} = f(z)^{-1}$ on $U_1$ which shows that $\lvert f \rvert > 1$ on $U_2 = \sigma U_1$.
However, the reflection may not be well-defined on all of $D_1$ or, if it is, $\sigma D_1$ may not be a superset of $D_2$, which leaves room for counter examples to your assumption.
The following does not satisfy all your requirements ($f$ has no essential singularity) but it shows what can happen. Let $\Delta$ be the open unit disc and $D = \Delta \setminus (-\mathrm i, 0]$ and $\Gamma = \{z \in D \mid \lvert z + \mathrm i \rvert^2 = 2 \}$. Let $f(z) = \tfrac12(z + z^{-1})$ on $D$. Then $\lvert f(z) \rvert < 1$ “above” $\Gamma$ and $\lvert f(z) \rvert = 1$ on $\Gamma$ but also $\lvert f(z) \rvert < 1$ “below” $\overline{\Gamma}$.