Since $F$ can be any continuous function, $G(s)=\mathrm e^{ks}F(s)$ can also be any continuous function. So we want
$\sup\limits_{x\in[0,r]}\left\{\mathrm e^{kx}\int\limits_0^x\mathrm e^{-ks}|G(s)|\,\,\,\mathrm ds\right\}\leq {c\over M}\sup\limits_{s\in[0,r]}\left\{|G(s)|\right\}$
for all continuous functions $G$. Given the supremum on the right-hand side, the left-hand side is maximal for constant $G$, so we can assume constant $G$ without loss of generality. The constant cancels, and thus we want
$ \begin{eqnarray} \sup\limits_{x\in[0,r]}\left\{\mathrm e^{kx}\int\limits_0^x\mathrm e^{-ks}\,\,\,\mathrm ds\right\} &=& \sup\limits_{x\in[0,r]}\left\{\mathrm e^{kx}\frac{1-\mathrm e^{-kx}}{k}\right\} \\ &=& \sup\limits_{x\in[0,r]}\left\{\frac{\mathrm e^{kx}-1}{k}\right\} \\ &\leq& {c\over M}\;. \end{eqnarray} $
(We can treat the case $k=0$ as a limiting case in which the argument of the supremum becomes simply $x$.) The argument of the supremum monotonically increases with $x$ independent of $k$, so the supremum is attained at $x=r$, and we get
$\frac{\mathrm e^{kr}-1}{k}\leq {c\over M}\;.$
The left-hand side monotonically increases with $k$, so you can find $c\lt1$ such that the inequality holds if $k\le k_{\text{max}}$, with $k_{\text{max}}$ determined by the transcendental equation
$\frac{\mathrm e^{k_{\text{max}}r}-1}{k_{\text{max}}}= {1\over M}\;.$