Question
Let $F(s)$ be a cumulative distribution function (cdf) of a random variable on $[0,\infty)$ which only has an atom at $0$, i.e. $F(0) >0$ and for all $s>0$:$$ \lim_{h\rightarrow 0}F(s+h)=F(s).$$ Let $\bar{F}(s)=1-F(s)$ and $f = F'(s)$. Note that$$ F(0) = 1-\int_0^\infty f(s) ds.$$
Now define the following operation on $F$: $$ HF(s)=\int_0^s \int_0^u (F(u)-F(u-v)) f(u-v) e^{-v}\, dv\, du. $$ Now suppose $F_1,F_2$ are both cdfs as described above, and suppose for all $s$, $F_1(s) \leq F_2(s)$ and $F_1(0)=F_2(0)$, show that $HF_1(s) \leq HF_2(s)$ for all $s$.
Thoughts
One can use Fubini to change the order of integration, the second part can then be solved by writing it as: \begin{align*} -\int_0^s \int_v^s F(u-v) f(u-v) \, du e^{-v}\, dv &= -\int_0^s\int_0^{s-v}F(u)f(u)\,du e^{-v}\,dv\\ &=-\frac{1}{2} \int_0^s \left( F(s-v)^2-F(0)^2 \right) e^{-v}\, dv. \end{align*} However we can not do something similar for the first part as here the argument of $F$ and $f$ are not the same..
Partal Integration
As suggested in the answer, partial integration can be used to get rid of the density $f$. An alternative partial integration to use to simplify the first part is to note that the first part equals: $$ \int_0^sF(u)\int_0^uf(u-v) e^{-v}dv du $$ and then use partial integration on $\int_0^uf(u-v) e^{-v}dv$, integrating $f(u-v)$ and differentiating $e^{-v}$.
This isn't a complete answer but it's too long for a comment. You can 'solve' the remaining part of the integral using partial integration as follows:
$$ \begin{align} \def\d{\,\mathrm{d}} \int_0^s \int_0^u F(u) f(u-v) e^{-v} \d v \d u &= \int_0^s \int_0^u F(u) f(v) e^{-(u-v)} \d v \d u \\ &= \int_0^s \int_0^u F(u) e^{-u} f(v) e^v\d v \d u \\ &= \int_0^s F(u) e^{-u} \left(F(u)e^u - F(0) - \int_0^u F(v) e^v \d v \right)\, \d u \\ &= \int_0^s F(u) \left(F(u) - F(0)e^{-u} - \int_0^u F(v) e^{-(u-v)} \d v \right)\, \d u \\ \end{align} $$
Combined with the answer you found for the other half this might be enough to prove the full inequality.