If $\lambda>0$ and $\lambda>\epsilon$ one has
$$\lim_{t\rightarrow\infty}e^{\epsilon t}\mathcal{S}=\int_0^\infty J_1\left(-Ae^{-\lambda s}\right)e^{\epsilon s}ds$$
$$\qquad=\frac{A/2}{\epsilon-\lambda} \, _1F_2\left(\frac{1}{2}-\frac{\epsilon}{2 \lambda};2,\frac{3}{2}-\frac{\epsilon}{2 \lambda};-A^2/4\right).$$
The asymptotics is different for other ranges of $\lambda,\epsilon$, for example, for $\lambda,\epsilon<0$ one has ${\cal S}\rightarrow -e^{-\epsilon t}J_1(Ae^{-\lambda t})/\epsilon$.
$\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$Take any $\mu\in\M(\Om)$, any open subset $\Th$ of $\Om$, and any real $\ep>0$. Let $\de:=\ep/4$.
By the Hahn decomposition theorem, there is a partition of $\Om$ into Borel sets $D^\pm$ such that $D^+$ is a positive set for $\mu$ and $D^-$ is a negative set for $\mu$.
Let
\begin{equation*}
A^\pm:=\Th\cap D^\pm. \tag{1}\label{1}
\end{equation*}
Since $|\mu|$ is inner regular, there exist compact sets
\begin{equation*}
K^\pm\subseteq A^\pm\text{ such that }|\mu|(A^\pm\setminus K^\pm)<\de. \tag{2}\label{2}
\end{equation*}
Since $\Om$ is normal, there exist open subsets $U^\pm$ of $\Th$ such that
\begin{equation*}
U^\pm\supseteq K^\pm\text{ and }U^+\cap U^-=\emptyset. \tag{3}\label{3}
\end{equation*}
Since the sets $K^\pm$ are compact and $\Om$ is locally compact, without loss of generality the closures of the sets $U^\pm$ are compact.
By Urysohn'slemma, there exist continuous functions $f^\pm\colon\Om\to\R$ such that
\begin{equation*}
0\le f^\pm\le1,\quad f^\pm=1\text{ on }K^\pm,\quad f^\pm=0\text{ on }\Om\setminus U^\pm. \tag{4}\label{4}
\end{equation*}
Let
\begin{equation*}
f:=f^+-f^-.
\end{equation*}
Then $f^+f^-=0$, whence $|f|\le1$. Also, $f=0$ on $\Om\setminus(U^+\cup U^-)$. So, recalling that the closures of the sets $U^\pm$ are compact, we see that $f\in C_c(\Om)$. Also, since $U^\pm$ are subsets of $\Th$, we have $|f|\le1_\Th$.
It remains to show that
\begin{equation*}
\int_\Om f\,d\mu\ge|\mu|(\Th)-\ep. \tag{$*$}\label{*}
\end{equation*}
To do this, note that, by \eqref{3}, \eqref{2}, and \eqref{1},
\begin{equation}
\begin{aligned}
|\mu|(U^-\setminus K^-)&\le|\mu|(\Th\setminus U^+\setminus K^-) \\
&=|\mu|(\Th)-|\mu|(U^+)-|\mu|(K^-) \\
&\le|\mu|(\Th)-|\mu|(K^+)-|\mu|(K^-) \\
&<|\mu|(\Th)-|\mu|(A^+)-|\mu|(A^-)+2\de=2\de.
\end{aligned}
\tag{5}\label{5}
\end{equation}
So, by \eqref{4}, \eqref{3}, \eqref{2}, \eqref{5}, and \eqref{1},
\begin{equation*}
\begin{aligned}
\int_\Om f\,d\mu&=\int_{U^+} f^+\,d\mu-\int_{U^-} f^-\,d\mu \\
&\ge\int_{K^+} f^+\,d\mu-\int_{K^-} f^-\,d\mu -\int_{U^-\setminus K^-} f^-\,d\mu \\
&=\mu(K^+)-\mu(K^-) -\int_{U^-\setminus K^-} f^-\,d\mu \\
&\ge\mu(K^+)-\mu(K^-) -|\mu|(U^-\setminus K^-) \\
&>\mu(A^+)-\de-\mu(A^-)-\de -2\de \\
&=|\mu|(\Th)-4\de=|\mu|(\Th)-\ep,
\end{aligned}
\end{equation*}
so that \eqref{*} is proved. $\quad\Box$
Best Answer
$\newcommand\EE{\mathcal E}\newcommand\la\lambda\newcommand\R{\mathbb R}\newcommand\ep\varepsilon$What you wanted us to prove is not true.
Indeed, take any $\phi\in\EE$ such that $\phi\ge1_{[-1/2,1/2]}$. Write $A\gg B$ for $A\ge cB$, where $c$ is a universal positive real constant.
Then, for $w:=x-y_2$,
\begin{equation} \begin{aligned} &\int_\R \phi_x^\la(y_1)p(r,y_1-y_2)\,dy_1 \\ &\gg\frac1\la\,\int_\R dy_1\, 1(|y_1-x|\le\la/2) \frac1{\sqrt r}\,\exp-\frac{(y_1-y_2)^2}{4r} \\ &=\frac1\la\,\int_\R dz\, 1(|z|\le\la/2) \frac1{\sqrt r}\,\exp-\frac{(w+z)^2}{4r} \\ &\ge\frac1\la\,\int_\R dz\, 1(|z|\le\la/2) \frac1{\sqrt r}\,\exp-\frac{w^2+z^2}{2r} \\ &\ge\exp\Big(-\frac{\la^2}{8r}\Big)\frac1{\sqrt r}\,\exp-\frac{w^2}{2r}. \end{aligned} \end{equation}
So, \begin{equation} \begin{aligned} &\int_\R dy_2\,\Big(\int_\R \phi_x^\la(y_1)p(r,y_1-y_2)\,dy_1\Big)^2 \\ &\gg \exp\Big(-\frac{\la^2}{4r}\Big) \int_\R dw\,\frac1r\,\exp-\frac{w^2}r \\ &\gg \frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{4r}\Big) \end{aligned} \end{equation} and hence \begin{equation} \begin{aligned} I&:=\int_0^{|v-u|}dr\,\int_\R dy_2\,\Big(\int_\R \phi_x^\la(y_1)p(r,y_1-y_2)\,dy_1\Big)^2 \\ &\gg \int_0^{|v-u|}dr\,\frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{4r}\Big) \\ &\ge \int_{|v-u|/2}^{|v-u|}dr\,\frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{4r}\Big) \\ &\ge \int_{|v-u|/2}^{|v-u|}dr\,\frac1{\sqrt r}\,\exp\Big(-\frac{\la^2}{2|v-u|}\Big) \\ &\gg |v-u|^{1/2}\exp\Big(-\frac{\la^2}{2|v-u|}\Big). \end{aligned} \end{equation}
Letting now, for instance, $U=1$, $v=1$, and $u=0$, for all $\la\in(0,1]$ we get \begin{equation} I\gg1. \end{equation} So, if $\beta<1/2$, then there is no real $\ep>0$ and $C>0$ such that $I^{1/2}\le C|v-u|^{\ep} \la^{1/2-\beta}$ for all $\la\in(0,1]$. $\quad\Box$