Let $G$ be the heat kernal, i.e. for $0\le t<s$ and $x,y\in\mathbb R$
$$G(t,x;s,y):=\frac{1}{\sqrt{4\pi(s-t)}}\exp\left(-\frac{(y-x)^2}{4(s-t)}\right).$$
For $T>0$, let $\mathcal H_T:=\{h:[0,T]\to [0,1]:~ h,~h' \mbox{ are both continuous on } [0,T]\}$ be endowed with the norm
$$\|h\|_T := \max_{0\le t\le T}|h(t)| + \max_{0\le t\le T}|h'(t)|,\quad \forall h\in\mathcal H_T.$$
Define the map $F$ on $\mathcal H_T$ as follows: $F[h]=\big(F[h](s): 0\le s\le T\big)$ with
$$F[h](s):=\int_{-s}^{\infty}\left(\int_0^s G\big(A(u), -u;A(s),y\big)h'(u)\frac{\big(1+h(u)\big)^2}{\big(1+h\circ A(u)\big)^2} du\right)dy + \int_{-s}^{\infty}\left(\int_0^{\infty} G\big(0,x;A(s),y\big)\rho(x)dx\right)dy,$$
where $A: \mathbb R_+\to\mathbb R_+$ denotes the inverse of the function
$$\mathbb R_+\ni t\mapsto \int_0^t (1+h(r))^2dr\in \mathbb R_+$$
and $\rho: \mathbb R_+\to \mathbb R_+$ is a probability density, i.e.
$$\int_0^{\infty}\rho(x)dx =1.$$
Can we prove the existence of $T>0$ s.t. $F$ is a contraction w.r.t. $\|\cdot\|_T$? Any answer, comments or references are appreciated.
PS : I strongly believe the answer is yes, while I am not familiar with the inequalities related to the heat kernal, especially when estimating the derivative $F[h]'$. Bty, $\circ$ stands for the composition of functions.
PS 2 : A straightforward computation yields for all $h_1, h_2 \in\mathcal H_T$,
$$|A_1(t)-A_2(t)|\le 4t\max_{0\le r\le t}|h_1(r)-h_2(r)|,\quad \forall 0\le t\le T.$$
Best Answer
$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\De}{\Delta}\newcommand\R{\mathbb R}$Edit: This answer is insufficient, even though (almost) all the reasoning appears relevant to the problem. I will try to come back and fix it.
Suppose that $T\in(0,1]$.
Integrating in $y$, simplify the expression for $F$ as follows: \begin{equation*} F[h](s)=I_1(h)+I_2(h), \tag{1} \end{equation*} where \begin{equation*} I_1(h):=\int_0^s \frac{1}{2} \Big(\text{erf}\Big(\frac{s-u}{2 \sqrt{A_h(s)-A_h(u)}}\Big)+1\Big) h'(u)\frac{\big(1+h(u)\big)^2}{\big(1+h\circ A_h(u)\big)^2}\, du, \end{equation*} \begin{equation*} I_2(h):=\int_0^{\infty} \frac{1}{2} \Big(\text{erf}\Big(\frac{s+x}{2 \sqrt{A_h(s)}}\Big)+1\Big)\rho(x)\,dx. \end{equation*} Here the notation $A_h$ reminds us that $A$ depends on $h$.
Let $H_T:=\mathcal H_T$. Let $B_T$ be the set of all bounded real-valued functions on $[0,T]$, endowed with the sup-norm.
Since $0\le h\le1$, the operator that maps any $h\in H_T$ to the function $[0,T]\ni u\mapsto\big(1+h(u)\big)^2$ in $L_T$ is Lipschitz with a universal Lipschitz constant.
Take any $h_0,h_1$ in $H_T$ with \begin{equation*} \de:=\|h_1-h_0\|<1/16. \tag{2} \end{equation*} Letting \begin{equation*} J_h(t):=\int_0^t (1+h(r))^2\,dr, \end{equation*} we see that $J_h$ is $4$-Lipschitz, $t\le J_h(t)\le4t$, $A_h$ is $1$-Lipschitz, $|J_{h_1}(t) -J_{h_0}(t)|\le8t\de$, and hence \begin{equation*} \frac u4\le A_{h_j}(u)\le\frac{A_{h_{1-j}(u)}}{1-8\de}\le\frac u{1-8\de}, \tag{3} \end{equation*} \begin{equation*} |A_{h_1}(u)-A_{h_0}(u)|\le\frac{8\de u}{1-8\de}\le16\de u; \tag{4} \end{equation*} here in what follows, $0\le t\le T$, $0<u<s\le T$, and $j\in\{0,1\}$, unless specified otherwise.
For brevity, let us refer to the Lipschitz maps with Lipschitz constants depending only on $\de$ as good-Lipschitz. In particular, the map $H_T\ni h\mapsto A_h$ is good-Lipschitz. Therefore, the maps $H_T\ni h\mapsto\big(1+h\circ A_h(u)\big)^2$ and \begin{equation*} H_T\ni h\mapsto h'(u)\frac{\big(1+h(u)\big)^2}{\big(1+h\circ A_h(u)\big)^2} \tag{5} \end{equation*} are good-Lipschitz, for each $u\in[0,T]$. The latter map is also bounded by a constant depending only on $\de$; let us refer to such maps as well-bounded.
For $h_t:=h_0+t(h_1-h_0)$, $\tau_t:=A_{h_t}(s)-A_{h_t}(u)$, and $\De\tau:=\tau_1-\tau_0$, we have $0<\frac{s-u}4\le\tau_t\le s-u$ (since $J_h$ is $4$-Lipschitz and $A_h$ is $1$-Lipschitz), $|\De\tau|\le32\de s$ (by (4)), and hence
the absolute value of the derivative of the map \begin{equation*} [0,1]\ni t\mapsto \dfrac{1}{2} \Big(\text{erf}\Big(\dfrac{s-u}{2 \sqrt{A_{h_t}(s)-A_{h_t}(u)}}\Big)+1\Big) \end{equation*} is \begin{equation*} \begin{aligned} & \frac{ s-u }{4 \sqrt{\pi } \tau_t^{3/2}}\ \exp\Big\{-\frac{(s-u)^2}{4\tau_t}\Big\}\,|\De\tau| \\ & \le\frac1{4^{-1/2} \sqrt{\pi } (s-u)^{1/2}}\ \exp\Big\{-\frac{s-u}4\Big\}\,32\de s. \end{aligned} \end{equation*} The integral in $u\in(0,s)$ of the latter expression is $\le CT$; here and in what follows, $C$ will denote various well-bounded expressions.
Therefore and because $|\text{erf}|\le1$ and the map (5) is good-Lipschitz and well-bounded, we conclude that $I_1$ is $CT$-Lipschitz.
It is similar but easier to show that $I_2$ is $CT$-Lipschitz as well. Indeed, still with $h_t:=h_0+t(h_1-h_0)$, let $\rho_t:=A_{h_t}(s)$, and $\De\rho:=\rho_1-\rho_0$. Then $0<\frac s4\le\rho_t\le s$ (since $J_h$ is $4$-Lipschitz and $A_h$ is $1$-Lipschitz), $|\De\rho|\le16\de s$ (by (4)), and hence
the absolute value of the derivative of the map \begin{equation*} [0,1]\ni t\mapsto \frac{1}{2} \Big(\text{erf}\Big(\frac{s+x}{2 \sqrt{A_{h_t}(s)}}\Big)+1\Big) \end{equation*} is \begin{equation*} \begin{aligned} &\frac{|s+x|}{4 \sqrt{\pi } \rho_t^{3/2}}\,\exp\Big\{-\frac{(s+x)^2}{4 \rho_t}\Big\}\,|\De\rho| \\ &=\frac{|s+x|}{4 \sqrt{\pi } \rho_t^{1/2}}\,\exp\Big\{-\frac{(s+x)^2}{4 \rho_t}\Big\}\, \frac{|\De\rho|}{\rho_t} \\ &\le\frac{|s+x|}{4 \sqrt{\pi } \rho_t^{1/2}}\,\exp\Big\{-\frac{(s+x)^2}{4 \rho_t}\Big\}\, 64\de\le C; \end{aligned} \end{equation*} so, the integral in $u\in(0,s)$ of the latter expression is $\le CT$. So, $I_2$ is $CT$-Lipschitz as well.
We conclude that $F$ is $CT$-Lipschitz and hence a contraction if $T$ is small enough.