= Solution
Since
$$
f'(z)=\pi\sinh(\pi z),
$$
the critical points are exactly
$$
\boxed{z=ki\quad(k\in\mathbb Z)}.
$$
The <holomorphic inverse function theorem> shows that $f$ is conformal locally everywhere else.
For $z=x+iy$,
$$
f(z)=\cosh(\pi x)\cos(\pi y)
+i\sinh(\pi x)\sin(\pi y).
$$
Its imaginary part is positive throughout $S$. On the three boundary pieces,
$$
\begin{aligned}
L_1&:\quad f(x)=\cosh(\pi x)\in(1,\infty),\\
L_2&:\quad f(iy)=\cos(\pi y)\in(-1,1),\\
L_3&:\quad f(x+i)=-\cosh(\pi x)\in(-\infty,-1).
\end{aligned}
$$
These intervals traverse the boundary of the upper half-plane. The inverse branch
$$
z=\frac1\pi\operatorname{arcosh}\zeta,
\qquad \Re z>0,\quad0<\Im z<1,
$$
exists for $\Im\zeta>0$, proving that the image is precisely
$$
\boxed{f(S)=\{\zeta:\Im\zeta>0\}}.
$$
This is the <hyperbolic-cosine half-strip map>.
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