Let's try to prove 
$$
(\star)\,\,\,\,\,\arg \frac{1-zf(s-u)}{1-zf(s+u)}< \pi/2- \arg(1-z\bar{z}f(2u)).
$$
Summing this up with analogous inequality 
$$
\arg \frac{1-zf(t+u)}{1-zf(t-u)}< \pi/2- \arg(1-z\bar{z}f(-2u))
$$
we get what we need. Denote $zf(s-u)=A$, $zf(s+u)=B$, $z\bar{z}f(2u)=C$. Then for functions $\varphi_1(x)=e^{isx}$, $\varphi_2(x)=\bar{z}e^{iux}$, $\varphi_3(x)=\bar{z}e^{-iux}$ in $(L^2,\mu)$ we have $A=\langle\varphi_1,\varphi_2\rangle$, $B=\langle\varphi_1,\varphi_3\rangle$, $C=\langle\varphi_2,\varphi_3\rangle$. Thus the following matrix is non-negative definite as the Gram matrix of our functions
$$
\pmatrix{\|\varphi_1\|^2&A&B\\\bar{A}&\|\varphi_2\|^2&C\\\bar{B}&\bar{C}&\|\varphi_3\|^2}.
$$
We may increase diagonal elements upto 1, of course, it remains non-negative definite. Thus the following matrix is non-negative-definite
$$
\pmatrix{1&A&B\\\bar{A}&1&C\\\bar{B}&\bar{C}&1}.
$$
And what we have to prove is that $\arg \frac{1-A}{1-B}<\pi/2-\arg(1-C)$, or, in other words, $\Re \frac{(1-A)(1-C)}{1-B}\geq 0$. This should not be hard if true, but it should be true if the original inequality holds (since when $z$ approaches 1 by appropriate way and $t=-s$ in initial equation we get the same statement, right)?