Technically, the answer is yes $u(s)=1$.
However, for a quite boring reason (as hinted at in the comment of Xogn Ambandl):
If $\overline{f}(a-s)=f(s)$, \overline{f(a-s)}=f(s)$, then $2 \ \Re{f}(s)$ would be holomorphic in that region, as it is equal to $f(s)+\overline{f}(s)= f(s)+\overline{f(s)}= f(s) + f(a-s)$ a sum of holomorphic functions. Yet, then as a real-valued holomorphic function $\Re{f}(s)$ is constant. And, so $f(s)$ is constant.
Thus, the only functions $f$ fulfilling your assumptions are constant functions, for which what you ask about is clear.
