Ok, I give it another try:
Proposition.
Let \( S=\{z:0<\Re(z)\le 1\} \), \( S_\epsilon=\{z:0<\Re(z)<1+\epsilon\} \), \( \epsilon>0 \) and \( D\supseteq S_\epsilon \) being a domain (open and connected) of definition and \( H\supseteq S_\epsilon \) being a domain (open and connected) of values for a holomorphic function.
Let \( f \) be a holomorphic function on \( D \), \( G:=f(S_\epsilon)\subseteq H \), such that
(0) \( H\subseteq f(D) \).
(1) \( f(0)=1 \)
(2) \( f(z+1)=F(f(z)) \)
(U) \( f^{-1}(G) \) has bounded real part.
Then \( g=f \) for every other on \( D \) holomorphic \( g \) satisfying (0), (1), (2), (U).
Proof.
\( h(z):=g^{-1}(z)-f^{-1}(z) \) has bounded real part on \( G \). We consider \( f^{-1} \) and \( g^{-1} \) and so \( h \) to be holomorphic on the same Riemann surface \( G \). \( \delta(z):=h(f(z))=g^{-1}(f(z))-z \) is a 1-periodic function, holomorphic on \( S_\epsilon \). As \( S_\epsilon\supset S \) it can be continued to an entire function, so it has to take on every complex value with at most one exception already on the strip \( S \) otherwise it is a constant. Now \( \delta(S)=\delta(S_\epsilon)=h(G) \) has bounded real part and hence can not take on every value, so \( h(z)=0 \) and \( g=f \).
Proposition.
Let \( S=\{z:0<\Re(z)\le 1\} \), \( S_\epsilon=\{z:0<\Re(z)<1+\epsilon\} \), \( \epsilon>0 \) and \( D\supseteq S_\epsilon \) being a domain (open and connected) of definition and \( H\supseteq S_\epsilon \) being a domain (open and connected) of values for a holomorphic function.
Let \( f \) be a holomorphic function on \( D \), \( G:=f(S_\epsilon)\subseteq H \), such that
(0) \( H\subseteq f(D) \).
(1) \( f(0)=1 \)
(2) \( f(z+1)=F(f(z)) \)
(U) \( f^{-1}(G) \) has bounded real part.
Then \( g=f \) for every other on \( D \) holomorphic \( g \) satisfying (0), (1), (2), (U).
Proof.
\( h(z):=g^{-1}(z)-f^{-1}(z) \) has bounded real part on \( G \). We consider \( f^{-1} \) and \( g^{-1} \) and so \( h \) to be holomorphic on the same Riemann surface \( G \). \( \delta(z):=h(f(z))=g^{-1}(f(z))-z \) is a 1-periodic function, holomorphic on \( S_\epsilon \). As \( S_\epsilon\supset S \) it can be continued to an entire function, so it has to take on every complex value with at most one exception already on the strip \( S \) otherwise it is a constant. Now \( \delta(S)=\delta(S_\epsilon)=h(G) \) has bounded real part and hence can not take on every value, so \( h(z)=0 \) and \( g=f \).