I just see that we can essentially simplify our conditions:
Proposition. Let \( S \) be a vertical strip somewhat wider than \( 1 \), i.e. \( S=\{z\in\mathbb{C}: x_1-\epsilon<\Re(z)<x_1+1\} \) for some \( x_1\in\mathbb{R} \) and \( \epsilon>0 \).
Let \( D \), \( G \), \( G' \) be three domains (open and connected) such that \( S\subseteq D \), \( G\subseteq G' \) and let \( F \) be holomorphic on \( G' \), let \( y_1\in G \). Then there exist at most one function \( f \) that satisifies
(1) \( f \) is holomorphic on \( D \) and \( f(S)\subseteq G\subseteq f(D)=G' \)
(2) \( f(z+1)=F(f(z)) \) for all \( z\in D \) and \( f(x_1)=y_1 \)
(3) There is a \( D'\subseteq D \) such that \( f: D' \leftrightarrow G \) is biholomorphic.
Proof. Let \( g,h \) be two function that satisfy the above conditions. Then the function \( \delta(z)=g^{-1}(h(z)) \) is holomorphic on \( S \) (because \( h(S)\subseteq G \) and (3)) and satisfies \( h(z)=g(\delta(z)) \). By (3) and (4)
\( \delta(z+1)=g^{-1}(F(h(z)))=g^{-1}(F(g(\delta(z))))=g^{-1}(g(\delta(z)+1))=\delta(z)+1 \) and \( \delta(0)=0 \).
So \( \delta \) can be continued from \( S \) to an entire function.
But the same is also true for \( \delta_2=h^{-1}(g(z)) \) by the same reasoning. But as \( \delta\circ\delta_2=g^{-1}\circ h\circ h^{-1}\circ g =\text{id} \) and \( \delta_2\circ \delta = \text{id} \), we see that \( \delta:\mathbb{C}\leftrightarrow\mathbb{C} \) is a bijection!
Again with Picard's big theorem we conclude that \( \delta(x)=x+c \) with \( c=0 \).\( \boxdot \)
Proposition. Let \( S \) be a vertical strip somewhat wider than \( 1 \), i.e. \( S=\{z\in\mathbb{C}: x_1-\epsilon<\Re(z)<x_1+1\} \) for some \( x_1\in\mathbb{R} \) and \( \epsilon>0 \).
Let \( D \), \( G \), \( G' \) be three domains (open and connected) such that \( S\subseteq D \), \( G\subseteq G' \) and let \( F \) be holomorphic on \( G' \), let \( y_1\in G \). Then there exist at most one function \( f \) that satisifies
(1) \( f \) is holomorphic on \( D \) and \( f(S)\subseteq G\subseteq f(D)=G' \)
(2) \( f(z+1)=F(f(z)) \) for all \( z\in D \) and \( f(x_1)=y_1 \)
(3) There is a \( D'\subseteq D \) such that \( f: D' \leftrightarrow G \) is biholomorphic.
Proof. Let \( g,h \) be two function that satisfy the above conditions. Then the function \( \delta(z)=g^{-1}(h(z)) \) is holomorphic on \( S \) (because \( h(S)\subseteq G \) and (3)) and satisfies \( h(z)=g(\delta(z)) \). By (3) and (4)
\( \delta(z+1)=g^{-1}(F(h(z)))=g^{-1}(F(g(\delta(z))))=g^{-1}(g(\delta(z)+1))=\delta(z)+1 \) and \( \delta(0)=0 \).
So \( \delta \) can be continued from \( S \) to an entire function.
But the same is also true for \( \delta_2=h^{-1}(g(z)) \) by the same reasoning. But as \( \delta\circ\delta_2=g^{-1}\circ h\circ h^{-1}\circ g =\text{id} \) and \( \delta_2\circ \delta = \text{id} \), we see that \( \delta:\mathbb{C}\leftrightarrow\mathbb{C} \) is a bijection!
Again with Picard's big theorem we conclude that \( \delta(x)=x+c \) with \( c=0 \).\( \boxdot \)