I want to create a sigmoidal curve $f(x)$ with the parameters $s$ and $\epsilon$ so that it has the following features:
$f(0) = 0 +\epsilon$
$f(s) = 1 - \epsilon$
$f'(s/2)=1$
Is this possible? If not, how can I design a curve with only the first two features?
This is one posibility: $$ f(x)=\frac{e^{ax}-1}{e^{ax}+e^{as/2}}. $$ It is clear that $0\le f(x)<1$, $f(0)=0$ and $\lim_{x\to\infty}f(x)=1$. Moreover $$ f'(x)=\frac{b \bigl(e^{\frac{b s}{2}}+1\bigr) e^{b x}}{\bigl(e^{\frac{b s}{2}}+e^{b x}\bigr)^2}>0\quad\text{and}\quad f''(s/2)=0. $$ Finally $$ 2<b<4\quad\text{and}\quad s=\frac2b\,\log\Bigl(\frac{2}{4-b}\Bigr)\implies f'(s/2)=1. $$ In the above I have assumed that you want the inflection point at $s/2$. Playing with the parameter $b$ you can make $f(x)$ small near $x=0$ and cloase to $1$ near $\infty$.
More examples can be obtained with the same idea looking for $f$ of the form $$ f(x)=\frac{g(x)}{c+g(x)} $$ where $g(0)=0$, $\lim_{x\to\infty}g(x)=\infty$ and $c>0$.