Using the substitution $t=s\sqrt y$, we have
\begin{equation*}
f(y,L)=\sqrt y\,g(L/\sqrt y,\sqrt y),
\end{equation*}
where
\begin{equation*}
g(r,x):=\int_r^\infty e^{-x(s+1/s)}\,ds,
\end{equation*}
$r>0$, $x>0$.
So, it is enough to study $g(r,x)$. Using now the substitution $t=s+1/s$ and letting
\begin{equation*}
F(t):=\frac t{\sqrt{t^2-4}}\quad\text{and}\quad t_r:=r+1/r,
\end{equation*}
we have
\begin{equation*}
g(r,x)=
\left\{
\begin{alignedat}{2}
&\frac12\,\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt &&\quad\text{if }r\ge1, \\
&\frac12\,\int_2^{t_r} e^{-xt}(F(t)-1)\,dt && \\
&+\frac12\,\int_2^\infty e^{-xt}(F(t)+1)\,dt
&&\quad\text{if }r\in(0,1].
\end{alignedat}
\right.
\tag{10}\label{10}
\end{equation*}
Integrating by parts twice, we get
\begin{equation*}
\begin{aligned}
I(r,x)&:=\int_{t_r}^\infty e^{-xt}(F(t)+1)\,dt \\
&=\frac1x\,e^{-xt_r}(F(t_r)+1)+\frac1x\,\int_{t_r}^\infty e^{-xt}F'(t)\,dt \\
&=\frac1x\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\
&+\frac1{x^2}\,\int_{t_r}^\infty e^{-xt}F''(t)\,dt.
\end{aligned}
\tag{20}\label{20}
\end{equation*}
Note now that for $t>2$
\begin{equation*}
F'(t)=-\frac{4}{\left(t^2-4\right)^{3/2}}<0,\quad F''(t)=\frac{12 t}{\left(t^2-4\right)^{5/2}}>0.
\end{equation*}
So, from our integration by parts we get
\begin{equation*}
2l(r,x)<I(r,x)<2u(r,x)
\end{equation*}
$r\ge1$ and $x>0$, where
\begin{equation*}
u(r,x):=\frac1{2x}\,e^{-xt_r}(F(t_r)+1)=\frac1{2x}e^{-x(r+1/r)}\frac{2r^2}{r^2-1}
\end{equation*}
and
\begin{equation*}
l(r,x):=\frac1{2x}\,e^{-xt_r}\Big(F(t_r)+1+\frac1x\,F'(t_r)\Big) \\
= \frac1{2x}e^{-x(r+1/r)}\Big(\frac{2r^2}{r^2-1}-\frac1x\frac{4 r^3}{\left(r^2-1\right)^3}\Big).
\end{equation*}
So, by \eqref{10},
\begin{equation*}
l(r,x)<g(r,x)<u(r,x) \quad\text{for }r>1.
\end{equation*}
In the original version of the OP, particular interest was expressed in large values of $y$ and $L$, so that $x=\sqrt y$ is large. Accordingly, we have $l(r,x)\sim u(r,x)$ and hence
\begin{equation*}
l(r,x)\sim g(r,x)\sim u(u,x)
\end{equation*}
as $x\to\infty$ uniformly in $r\ge r_0$, for each real $r_0>1$, which means that the upper and lower bounds $u(r,x)$ and $l(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.
Consider now the case $r\in(0,1)$. Then, by \eqref{10},
\begin{equation*}
2g(r,x)=I(r,x)+2J(r,x),
\end{equation*}
where $I(r,x)$ is as in \eqref{20} and
\begin{equation*}
J(r,x):=\int_2^{t_r} e^{-xt}F(t)\,dt=e^{-2x}\int_0^{z_r} e^{-xz}G(z)\,dz, \tag{30}\label{30}
\end{equation*}
where $z_r:=t_r-2=r+1/r-2=(1/\sqrt r-\sqrt r)^2$ and
\begin{equation*}
\frac1{\sqrt z}<G(z):=\frac1{\sqrt z}\frac{2+z}{\sqrt{4+z}}<\frac{1+3z/8}{\sqrt z} \tag{40}\label{40}
\end{equation*}
for real $z>0$.
Using these lower and upper bounds on $G(z)$ together with \eqref{30}, we get
\begin{equation*}
l_1(r,x)<J(r,x)<u_1(r,x),
\end{equation*}
where
\begin{equation*}
l_1(r,x):=e^{-2x}\frac{\sqrt{\pi }\, \text{erf}\left(\sqrt{x
z_r}\right)}{\sqrt{x}},
\end{equation*}
\begin{equation*}
u_1(r,x):=l_1(r,x)
+e^{-2x}\frac{3\sqrt{\pi }\, \text{erf}\left(\sqrt{x
z_r}\right)}{16x^{3/2}}
-e^{-2x}\frac{3 \sqrt{z_r} e^{-x
z_r}}{8x}.
\end{equation*}
So,
\begin{equation*}
L(r,x):=l(r,x)+l_1(r,x)<g(r,x)<U(r,x):=u(r,x)+u_1(r,x) \quad\text{for }r\in(0,1)
\end{equation*}
and
\begin{equation*}
L(r,x)\sim g(r,x)\sim U(u,x)
\end{equation*}
as $x\to\infty$ uniformly in $r\in(0,r_0]$, for each real $r_0\in(0,1)$, which means that the upper and lower bounds $U(r,x)$ and $L(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.
Using the inequality $G(z)<1+1/\sqrt z$ instead of the second inequality in \eqref{40}, we get
\begin{equation*}
J(r,x)<\tilde u_1(r,x):=l_1(r,x)+ e^{-2x}\frac{1-e^{-xz_r}}x.
\end{equation*}
So,
\begin{equation*}
L(r,x)<g(r,x)<\tilde U(r,x):=u(r,x)+\tilde u_1(r,x) \quad\text{for }r\in(0,1)
\end{equation*}
and
\begin{equation*}
L(r,x)\sim g(r,x)\sim \tilde U(u,x)
\end{equation*}
as $x\to\infty$ uniformly in $r\in(0,r_0]$, for each real $r_0\in(0,1)$, which means that the upper and lower bounds $\tilde U(r,x)$ and $L(r,x)$ on $g(r,x)$ are asymptotically exact for such $r$ and $x$.
The upper bound $\tilde U(r,x)$ on $g(r,x)$ is simpler than $U(r,x)$, but not as accurate for large $x$.
