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 $$ M=\pmatrix{1&A&B\\\bar{A}&1&C\\\bar{B}&\bar{C}&1}. $$ And what we have to prove is that $\arg (1-A)(1-\bar{B})=\arg \frac{1-A}{1-B}<\pi/2-\arg(1+C)$, or, in other words, $\Re (1-A)(1-\bar{B})(1+C)\geq 0$. If I am not mistaken we have $$ 2\Re (1-A)(1-\bar{B})(1+C)=|-C+\bar{A}+B-1|^2+\det M. $$
Well, this is a non-strict inequality, something else must be said using $|z|<1$.