As far as I can tell this involves little to no complex analysis. Also, I'm assuming you want $S$ to be open in the upper half plane, not necessarily in all of $\mathbb{C}$.
To construct $S$, consider any $[a,b] \times [0, R] \subseteq \mathbb{H}$ (that is, $\{z: a \leqslant Re z \leqslant b, 0 \leqslant Im z \leqslant R \}$, for $R >> 0$, by compactness and hence uniform continuity there exists a subset $S_{a,b}$ of the form $[a,b] \times [0, \epsilon)$ on which $f$ is nonvanishing.
Now taking $[a,b]$ to be say $\ldots [-1,0], [0,1], [1,2], \ldots$, and patching your $S_{a,b}$ together nicely at the endpoints gives $S$.
$S$ is clearly simply connected (it deformation retracts onto the real line via "squishing").
Lastly, to construct $\psi$, we have $\Omega := S - \mathbb{R}$ is open and simply connected, $f$ is nonvanishing on $\Omega$, and hence $f = e^{g(z)}$ for $g$ holomorphic. Take $\psi := g/i$.
If this is unclear (or wrong!) lemme know.
A Möbius transformation that maps the upper half plane to itself must map the boundary, i.e. the real line, to itself as well. With the standard definition for the transformation,
$$f(z) = \frac{az+b}{cz+d},$$
it is fairly straightforward to see that the parameters $a, b, c, d$ must be real to map the real line to itself (try plugging in $z=0$, $z=\infty$, and then one other real number).
Then since $f$ is supposed to interchange $z_1$ and $z_2$, that means applying $f$ twice should send $z_1$ back to itself, and the same for $z_2$. Note that $f(f(z))$ is again another Möbius transformation with real coefficients. You can now solve the equation for the fixed points of the transformation (or look up the formula). This should answer the question about whether the transformation is possible.
Best Answer
Regardless of the correctness of the sign in the question, the answer is no, meaning that there are analytic functions from the upper half plane onto itself which are not Möbius transformations. See for instance this answer or the book The Problem of Moments by Shohat and Tamarkin.