Here is an idea and I will leave details to you (it might be that I made a stupid mistake somewhere therefore these computations must be checked carefully). All functions below should be sufficiently nice so that all formulas make sense. First I will formulate a lemma.
Lemma:
Let $p>1$. If $\psi(0)=0, \psi' >0$ and $\psi(\infty)=\infty$ then for any $f\geq 0$ and any (nice) measures $d\mu(t)$ we have
$$
\int_{0}^{\infty}\left(\int_{0}^{t}f(s)ds\right)^{p}d\mu(t)\leq \int_{0}^{\infty}f(y)^{p} \left[\frac{1}{(\psi'(y))^{p-1}}\int_{y}^{\infty}\psi^{p-1}(t)d\mu(t) \right]dy \quad (1)
$$
Remarks:
1) We have a freedom of choosing $\psi$. One can optimize the quantity $\left[\frac{1}{(\psi'(y))^{p-1}}\int_{y}^{\infty}\psi^{p-1}(t)d\mu(t) \right]$ in the right hand side of (1) over all $\psi$ and find the best bounds (This becomes some ODE problem which should not be a difficult problem so that I will leave details to the readers).
However, for our purposes we do not need to know the best $\psi$. One just needs to guess the right one to find some bound. Let me show you how it works on your example. In your case $d\mu(t)=e^{-t}dt$. Instead of $\psi$ take something like $\psi(t)=te^{\frac{t}{10(p-1)}}$. Then $\left[\frac{1}{(\psi'(y))^{p-1}}\int_{y}^{\infty}\psi^{p-1}(t)d\mu(t) \right] \leq C e^{-y}$. So the claim follows.
Proof of Lemma:
Let $\psi$ be a convex function, and let $w$ be some positive function (we will choose them later). Then Jensen's inequality together with Fubini's theorem implies:
\begin{align*}
&\int_{0}^{\infty} \varphi\left( \frac{1}{x}\int_{0}^{x} g(s)ds\right) w(x)dx\leq \int_{0}^{\infty}\frac{w(x)}{x}\int_{0}^{x}\varphi(g(s))dsdx=\\
&\int_{0}^{\infty} \varphi(g(s))\int_{s}^{\infty}\frac{w(x)}{x} dxds
\end{align*}
Let $\varphi=u^{p}$ and lets make a change of variables $s=\psi(y)$ and $x=\psi(t)$. Then the above inequality takes the form:
\begin{align*}
\int_{0}^{\infty} \left(\int_{0}^{t} g(\psi(y))\psi'(y) ds\right)^{p} \frac{w(\psi(t))}{\psi(t)^{p}}\psi'(t)dt\leq
\int_{0}^{\infty} [g(\psi(y))\psi'(y)]^{p}\int_{y}^{\infty}\frac{w(\psi(t)) \psi'(t)}{\psi(t) (\psi'(y))^{p-1}} dtdy
\end{align*}
Now choose $g$ so that $g(\psi(y))\psi'(y)=f(y)$. And choose $\frac{w(\psi(t))}{\psi(t)^{p}}\psi'(t)dt = d\mu(t)$ (for example take $w(\psi(t))dt =\frac{\psi^{p}(t)d\mu(t)}{\psi'(t)}$ ). Then the lemma follows.