Is the function
$$ A:t\in [0,1]\longrightarrow A(t)=\int_0^t e^{s-t} f(s)\,ds\in\mathbb{R} $$ always continuous or do we need hyphothesis on $f$ such as $f$ Riemann-integrable?
I know that the fundamental theorem of calculus would imply the continuity if the integrand depends only on $s$, but I don't know any result in this case.
If the integral exists for every $t$ in question (i.e. you may consider this on bounded intervals) then $$A(t+h)-A(t) = \int_0^{t+h}e^{s-t-h}f(s) \, ds - \int_0^t e^{s-t}f(s) \,ds$$ Now $$\int_0^{t+h}e^{s-t-h}f(s) \, ds =e^{-h}\left( \int_0^{t}e^{s-t}f(s) \, ds + \int_t^{t+h}e^{s-t}f(s) \, ds\right ) \\ = e^{-h}\left( \int_0^{t}e^{s-t}f(s) \, ds + h b(t)\right )$$ where $b(t) $ is some bounded function, so $$|A(t+h) - A(t)| \le |(1-e^{-h})| \int_0^{t}e^{s-t}f(s) \, ds + |h||b(s)$$ This tends to $0$ if $h\rightarrow 0$ and $f$ is integrable, (e.g. Riemann integrable) so yes.