In the rewritten form it doesn't require any ingenuity at all. For brevity, introduce the notation $$ (Th)(s)=\frac 1\pi\int_0^s\frac{h(s-u)h(u)}{\sqrt{u(s-u)}}\,du=\frac 1\pi\int_0^1\frac{h(s(1-v))h(sv)}{\sqrt{v(1-v)}}\,dv $$ Suppose that $h_0$ is any real valued Lipschitz function on $[0,+\infty)$ (the finiteness of the second moment guarantees Lipschitzness for your $h_0$ and, if you really want to discuss it, I can show how this particular condition can be dropped) such that $\|h_0\|_\infty=h_0(0)=1$ and there exists $h_0'(0)=-c< 0$ (the latter two conditions are essential). Then the iterations $h_{n+1}=Th_n$ converge to $e^{-cx}$ uniformly on compact subsets of $[0,+\infty)$.

The proof consists of several easy observations:

1. $\|Th\|_\infty\le\|h\|_\infty^2$ and, thereby, $\|h_n\|_\infty\le 1$ for all $n$.

2. If $h$ is bounded and $L$-Lipschitz, then $Th$ is $L\|h\|_\infty$-Lipschitz.

Indeed, $$ Th(s)-Th(S)=\frac1\pi\int_0^1\frac{dv}{\sqrt{v(1-v)}}[h(s(1-v))h(sv)-h(S(1-v))h(Sv)] $$ and $$ |h(s(1-v))h(sv)-h(S(1-v))h(Sv)| \\ \le\|h\|_\infty[|h(s(1-v))-h(S(1-v))|+|h(sv)-h(Sv)|] \\ \le\|h\|_\infty L[|s-S|(1-v+v)]=\|h\|_\infty L|s-S|\,. $$ Thus all $h_n$ are Lipschitz with the same Lipschitz constant $L$ as $h_0$.

3. If $A\in\mathbb R$ and $h(s)\ge e^{-As}$ on $[0,s_0]$, then $(Th)(s)\ge e^{-As}$ on $[0,s_0]$. 3') If $a\in\mathbb R$ and $0\le h(s)\le e^{-as}$ on $[0,s_0]$, then $(Th)(s)\le e^{-as}$ on $[0,s_0]$.

(all exponential functions are fixed points of $T$ and we have monotonicity as long as $h$ stays non-negative)

4. Now choose any $0<b<a<c<A<B$ and choose $s_0>0$ so that $e^{-As}\le h_0(s)\le e^{-as}$ on $[0,s_0]$ (by the derivative at zero condition such $s_0$ exists).

Consider the largest $S_n$ such that $e^{-Bs}\le h_n(s)\le e^{-bs}$ on $[0,S_n]$. We have $S_0\ge s_0$ and $S_{n+1}\ge S_n$. We want to improve the latter trivial inequality to some quantitative advance $S_{n+1}\ge S_n+\delta(S_n)$ where $\delta>0$ is separated from $0$ on any compact subinterval of $[s_0,+\infty)$. That is quite easy: $$ h_{n+1}(S_n)=(Th_n)(S_n)=\frac1\pi\int_0^{S_n}\frac{h_n(S_n-u)h_n(u)}{\sqrt{u(S_n-u)}}\,du\ge \\ \frac1\pi\int_0^{S_n}\frac{e^{-B(S_n-u)}e^{-Bu}}{\sqrt{u(S_n-u)}}\,du +\frac1\pi\int_0^{s_0}\frac{e^{-B(S_n-u)}[e^{-bu}-e^{-Bu}]}{\sqrt{u(S_n-u)}}\,du \\ \ge e^{-BS_n}+e^{-BS_n}S_n^{-1}\int_0^{s_0}[e^{-bu}-e^{-Bu}]\,du=e^{-BS_n}+\Delta(S_n) $$ Then, by the Lipschitz property of both $h_{n+1}$ and $e^{-Bs}$, the inequality $h_{n+1}(s)\ge e^{-Bs}$ persists on $[S_n,S_n+\delta(S_n)]$ with $\delta(S_n)=\frac{\Delta(S_n)}{B+L}$, say. The extension of the upper bound is similar.

The outcome is that the double inequality $e^{-Bs}\le h_n(s)\le e^{-bs}$ propagates from $[0,s_0]$ to the entire real line. Since $0<b<a<c<A<B$ were arbitrary, we conclude that $h_n$ tend to $e^{-cs}$ uniformly on compact intervals, finishing the story.

In the rewritten form it doesn't require any ingenuity at all. For brevity, introduce the notation $$ (Th)(s)=\frac 1\pi\int_0^s\frac{h(s-u)h(u)}{\sqrt{u(s-u)}}\,du=\frac 1\pi\int_0^1\frac{h(s(1-v))h(sv)}{\sqrt{v(1-v)}}\,dv $$ Suppose that $h_0$ is any real valued Lipschitz function on $[0,+\infty)$ (the finiteness of the second moment guarantees Lipschitzness for your $h_0$ and, if you really want to discuss it, I can show how this particular condition can be dropped) such that $\|h_0\|_\infty=h_0(0)=1$ and there exists $h_0'(0)=-c< 0$ (the latter two conditions are essential). Then the iterations $h_{n+1}=Th_n$ converge to $e^{-cx}$ uniformly on compact subsets of $[0,+\infty)$.

The proof consists of several easy observations:

1. $\|Th\|_\infty\le\|h\|_\infty^2$ and, thereby, $\|h_n\|_\infty\le 1$ for all $n$.

2. If $h$ is bounded and $L$-Lipschitz, then $Th$ is $L\|h\|_\infty$-Lipschitz.

Indeed, $$ Th(s)-Th(S)=\frac1\pi\int_0^1\frac{dv}{\sqrt{v(1-v)}}[h(s(1-v))h(sv)-h(S(1-v))h(Sv)] $$ and $$ |h(s(1-v))h(sv)-h(S(1-v))h(Sv)| \\ \le\|h\|_\infty[|h(s(1-v))-h(S(1-v))|+|h(sv)-h(Sv)|] \\ \le\|h\|_\infty L[|s-S|(1-v+v)]=\|h\|_\infty L|s-S|\,. $$ Thus all $h_n$ are Lipschitz with the same Lipschitz constant $L$ as $h_0$.

3. If $A\in\mathbb R$ and $h(s)\ge e^{-As}$ on $[0,s_0]$, then $(Th)(s)\ge e^{-As}$ on $[0,s_0]$. 3') If $a\in\mathbb R$ and $0\le h(s)\le e^{-as}$ on $[0,s_0]$, then $(Th)(s)\le e^{-as}$ on $[0,s_0]$.

(all exponential functions are fixed points of $T$ and we have monotonicity as long as $h$ stays non-negative)

4. Now choose any $0<b<a<c<A<B$ and choose $s_0>0$ so that $e^{-As}\le h_0(s)\le e^{-as}$ on $[0,s_0]$ (by the derivative at zero condition such $s_0$ exists).

Consider the largest $S_n$ such that $e^{-Bs}\le h_n(s)\le e^{-bs}$ on $[0,S_n]$. We have $S_0\ge s_0$ and $S_{n+1}\ge S_n$. We want to improve the latter trivial inequality to some quantitative advance $S_{n+1}\ge S_n+\delta(S_n)$ where $\delta>0$ is separated from $0$ on any compact subinterval of $[s_0,+\infty)$. That is quite easy: $$ h_{n+1}(S_n)=(Th_n)(S_n)=\frac1\pi\int_0^{S_n}\frac{h_n(S_n-u)h_n(u)}{\sqrt{u(S_n-u)}}\,du\ge \\ \frac1\pi\int_0^{S_n}\frac{e^{-B(S_n-u)}e^{-Bu}}{\sqrt{u(S_n-u)}}\,du +\frac1\pi\int_0^{s_0}\frac{e^{-B(S_n-u)}[e^{-Au}-e^{-Bu}]}{\sqrt{u(S_n-u)}}\,du \\ \ge e^{-BS_n}+e^{-BS_n}S_n^{-1}\int_0^{s_0}[e^{-Au}-e^{-Bu}]\,du=e^{-BS_n}+\Delta(S_n) $$ Then, by the Lipschitz property of both $h_{n+1}$ and $e^{-Bs}$, the inequality $h_{n+1}(s)\ge e^{-Bs}$ persists on $[S_n,S_n+\delta(S_n)]$ with $\delta(S_n)=\frac{\Delta(S_n)}{B+L}$, say. The extension of the upper bound is similar.

The outcome is that the double inequality $e^{-Bs}\le h_n(s)\le e^{-bs}$ propagates from $[0,s_0]$ to the entire real line. Since $0<b<a<c<A<B$ were arbitrary, we conclude that $h_n$ tend to $e^{-cs}$ uniformly on compact intervals, finishing the story.

In the rewritten form it doesn't require any ingenuity at all. For brevity, introduce the notation $$ (Th)(s)=\frac 1\pi\int_0^s\frac{h(s-u)h(u)}{\sqrt{u(s-u)}}\,du=\frac 1\pi\int_0^1\frac{h(s(1-v))h(sv)}{\sqrt{v(1-v)}}\,dv $$ Suppose that $h_0$ is any real valued Lipschitz function on $[0,+\infty)$ (the finiteness of the second moment guarantees Lipschitzness for your $h_0$ and, if you really want to discuss it, I can show how this particular condition can be dropped) such that $\|h_0\|_\infty=h_0(0)=1$ and there exists $h_0'(0)=-c< 0$ (the latter two conditions are essential). Then the iterations $h_{n+1}=Th_n$ converge to $e^{-cx}$ uniformly on compact subsets of $[0,+\infty)$.

The proof consists of several easy observations:

1. $\|Th\|_\infty\le\|h\|_\infty^2$ and, thereby, $\|h_n\|_\infty\le 1$ for all $n$.

2. If $h$ is bounded and $L$-Lipschitz, then $Th$ is $L\|h\|_\infty$-Lipschitz.

Indeed, $$ Th(s)-Th(S)=\frac1\pi\int_0^1\frac{dv}{\sqrt{v(1-v)}}[h(s(1-v))h(sv)-h(S(1-v))h(Sv)] $$ and $$ |h(s(1-v))h(sv)-h(S(1-v))h(Sv)| \\ \le\|h\|_\infty[|h(s(1-v))-h(S(1-v))|+|h(sv)-h(Sv)|] \\ \le\|h\|_\infty L[|s-S|(1-v+v)]=\|h\|_\infty L|s-S|\,. $$ Thus all $h_n$ are Lipschitz with the same Lipschitz constant $L$ as $h_0$.

3. If $A\in\mathbb R$ and $h(s)\ge e^{-As}$ on $[0,s_0]$, then $(Th)(s)\ge e^{-As}$ on $[0,s_0]$. 3') If $a\in\mathbb R$ and $0\le h(s)\le e^{-as}$ on $[0,s_0]$, then $(Th)(s)\ge e^{-as}$ on $[0,s_0]$.

(all exponential functions are fixed points of $T$ and we have monotonicity as long as $h$ stays non-negative)

4. Now choose any $0<b<a<c<A<B$ and choose $s_0>0$ so that $e^{-As}\le h_0(s)\le e^{-as}$ on $[0,s_0]$ (by the derivative at zero condition such $s_0$ exists).

Consider the largest $S_n$ such that $e^{-Bs}\le h_n(s)\le e^{-bs}$ on $[0,S_n]$. We have $S_0\ge s_0$ and $S_{n+1}\ge S_n$. We want to improve the latter trivial inequality to some quantitative advance $S_{n+1}\ge S_n+\delta(S_n)$ where $\delta>0$ is separated from $0$ on any compact subinterval of $[s_0,+\infty)$. That is quite easy: $$ h_{n+1}(S_n)=(Th_n)(S_n)=\frac1\pi\int_0^{S_n}\frac{h(S_n-u)h(u)}{\sqrt{u(S_n-u)}}\,du\ge \\ \frac1\pi\int_0^{S_n}\frac{e^{-B(S_n-u)}e^{-Bu}}{\sqrt{u(S_n-u)}}\,du +\frac1\pi\int_0^{s_0}\frac{e^{-B(S_n-u)}[e^{-bu}-e^{-Bu}]}{\sqrt{u(S_n-u)}}\,du \\ \ge e^{-BS_n}+e^{-BS_n}S_n^{-1}\int_0^{s_0}[e^{-bu}-e^{-Bu}]\,du=e^{-BS_n}+\Delta(S_n) $$ Then, by the Lipschitz property of both $h_{n+1}$ and $e^{-Bs}$, the inequality $h_{n+1}(s)\ge e^{-Bs}$ persists on $[S_n,S_n+\delta(S_n)]$ with $\delta(S_n)=\frac{\Delta(S_n)}{B+L}$, say. The extension of the upper bound is similar.

The outcome is that the double inequality $e^{-Bs}\le h_n(s)\le e^{-bs}$ propagates from $[0,s_0]$ to the entire real line. Since $0<b<a<c<A<B$ were arbitrary, we conclude that $h_n$ tend to $e^{-cs}$ uniformly on compact intervals, finishing the story.

In the rewritten form it doesn't require any ingenuity at all. For brevity, introduce the notation $$ (Th)(s)=\frac 1\pi\int_0^s\frac{h(s-u)h(u)}{\sqrt{u(s-u)}}\,du=\frac 1\pi\int_0^1\frac{h(s(1-v))h(sv)}{\sqrt{v(1-v)}}\,dv $$ Suppose that $h_0$ is any real valued Lipschitz function on $[0,+\infty)$ (the finiteness of the second moment guarantees Lipschitzness for your $h_0$ and, if you really want to discuss it, I can show how this particular condition can be dropped) such that $\|h_0\|_\infty=h_0(0)=1$ and there exists $h_0'(0)=-c< 0$ (the latter two conditions are essential). Then the iterations $h_{n+1}=Th_n$ converge to $e^{-cx}$ uniformly on compact subsets of $[0,+\infty)$.

The proof consists of several easy observations:

1. $\|Th\|_\infty\le\|h\|_\infty^2$ and, thereby, $\|h_n\|_\infty\le 1$ for all $n$.

2. If $h$ is bounded and $L$-Lipschitz, then $Th$ is $L\|h\|_\infty$-Lipschitz.

Indeed, $$ Th(s)-Th(S)=\frac1\pi\int_0^1\frac{dv}{\sqrt{v(1-v)}}[h(s(1-v))h(sv)-h(S(1-v))h(Sv)] $$ and $$ |h(s(1-v))h(sv)-h(S(1-v))h(Sv)| \\ \le\|h\|_\infty[|h(s(1-v))-h(S(1-v))|+|h(sv)-h(Sv)|] \\ \le\|h\|_\infty L[|s-S|(1-v+v)]=\|h\|_\infty L|s-S|\,. $$ Thus all $h_n$ are Lipschitz with the same Lipschitz constant $L$ as $h_0$.

3. If $A\in\mathbb R$ and $h(s)\ge e^{-As}$ on $[0,s_0]$, then $(Th)(s)\ge e^{-As}$ on $[0,s_0]$. 3') If $a\in\mathbb R$ and $0\le h(s)\le e^{-as}$ on $[0,s_0]$, then $(Th)(s)\le e^{-as}$ on $[0,s_0]$.

(all exponential functions are fixed points of $T$ and we have monotonicity as long as $h$ stays non-negative)

4. Now choose any $0<b<a<c<A<B$ and choose $s_0>0$ so that $e^{-As}\le h_0(s)\le e^{-as}$ on $[0,s_0]$ (by the derivative at zero condition such $s_0$ exists).

Consider the largest $S_n$ such that $e^{-Bs}\le h_n(s)\le e^{-bs}$ on $[0,S_n]$. We have $S_0\ge s_0$ and $S_{n+1}\ge S_n$. We want to improve the latter trivial inequality to some quantitative advance $S_{n+1}\ge S_n+\delta(S_n)$ where $\delta>0$ is separated from $0$ on any compact subinterval of $[s_0,+\infty)$. That is quite easy: $$ h_{n+1}(S_n)=(Th_n)(S_n)=\frac1\pi\int_0^{S_n}\frac{h_n(S_n-u)h_n(u)}{\sqrt{u(S_n-u)}}\,du\ge \\ \frac1\pi\int_0^{S_n}\frac{e^{-B(S_n-u)}e^{-Bu}}{\sqrt{u(S_n-u)}}\,du +\frac1\pi\int_0^{s_0}\frac{e^{-B(S_n-u)}[e^{-bu}-e^{-Bu}]}{\sqrt{u(S_n-u)}}\,du \\ \ge e^{-BS_n}+e^{-BS_n}S_n^{-1}\int_0^{s_0}[e^{-bu}-e^{-Bu}]\,du=e^{-BS_n}+\Delta(S_n) $$ Then, by the Lipschitz property of both $h_{n+1}$ and $e^{-Bs}$, the inequality $h_{n+1}(s)\ge e^{-Bs}$ persists on $[S_n,S_n+\delta(S_n)]$ with $\delta(S_n)=\frac{\Delta(S_n)}{B+L}$, say. The extension of the upper bound is similar.

The outcome is that the double inequality $e^{-Bs}\le h_n(s)\le e^{-bs}$ propagates from $[0,s_0]$ to the entire real line. Since $0<b<a<c<A<B$ were arbitrary, we conclude that $h_n$ tend to $e^{-cs}$ uniformly on compact intervals, finishing the story.