From Theorem 0.3.1 of [1], we have that a kernel operator $$ T(f)\equiv \int_0^\infty K(x,x') f(x')dx',\;\;K(x,y)=e^{x-x'}\chi_{y>x>0} $$ is such that $$ \sup_{x'}\left(\int_0^\infty K(x,x') dx\right),\;\; \sup_{x}\left(\int_0^\infty K(x,x') dx'\right)\leq C. $$ Then $$ \|T(f)\|_{L^2(\mathbb{R}^+)}\leq C \|f\|_{L^2(\mathbb{R}^+)}. $$ Thus $T$ defines a continuous operator on $L^2(\mathbb{R}^+)$. In my specific case, $K(x,y)=e^{x-x'}\chi_{y>x>0}$, $C=1$ and thus that answers my question!

A big thank you to Christian Remling to point me toward that theorem.

[1] Sogge, Christopher D., Fourier integrals in classical analysis, Cambridge Tracts in Mathematics. 105. Cambridge: Cambridge University Press. x, 237 p. (1993). ZBL0783.35001.

From Theorem 0.3.1 of [1], we have that if a kernel operator $$ T(f)\equiv \int_0^\infty K(x,x') f(x')dx',\;\;K(x,y)=e^{x-x'}\chi_{y>x>0} $$ is such that $$ \sup_{x'}\left(\int_0^\infty K(x,x') dx\right),\;\; \sup_{x}\left(\int_0^\infty K(x,x') dx'\right)\leq C $$ then $$ \|T(f)\|_{L^2(\mathbb{R}^+)}\leq C \|f\|_{L^2(\mathbb{R}^+)}. $$ Thus $T$ defines a continuous operator on $L^2(\mathbb{R}^+)$. In my specific case, $K(x,y)=e^{x-x'}\chi_{y>x>0}$, $C=1$ and thus that answers my question!

A big thank you to Christian Remling to point me toward that theorem.

[1] Sogge, Christopher D., Fourier integrals in classical analysis, Cambridge Tracts in Mathematics. 105. Cambridge: Cambridge University Press. x, 237 p. (1993). ZBL0783.35001.