Let me summarize and condense what @IosefPinelis has said.

We are to estimate $$f(y,L) = \int_L^\infty e^{-t-y/t} dt = \sqrt{y} g(L/\sqrt{y},\sqrt{y}),$$
where $g(r,x)=\int_r^\infty e^{-x(s+1/s)} ds.$ By a substitution of variables $t = r+1/r$,
$$g(r,x) = \frac{1}{2} I(r,x) + \begin{cases} 0 &\text{if $r\geq 1$,}\\ J(r,x)& \text{if $0<r<1$,}\end{cases}$$
where $$I(r,x) = \int_{r+1/r}^\infty e^{-x t} (F(t)+1) dt,\;\;\;\;\;\;\;
J(r,x) = \int_2^{r+1/r} e^{-x t} F(t) dt$$ and $F(t) = \frac{t}{\sqrt{t^2-4}}$. (We are using the fact that $F(t)$ is an odd function.) Letting $t = u+2$, we see that
$$I(r,x) = e^{-2 x} \int_{u_r}^\infty e^{-x u} (G(u)+1) du,
\;\;\;\;J(r,x)= e^{-2 x}\int_0^{u_r} e^{-x u} G(u) du,$$
where $G(u) = F(u+2) = \frac{1}{\sqrt{u}} \frac{1+u/2}{\sqrt{1+u/4}}$ for $u\geq 0$ and $u_r = r+1/r-2$.

We can give different upper bounds on $G(u)$. @IosefPinelis used the bound $H(u)\leq H(u_r) + (u-u_r) H'(u_r)$ for $H(u) = \sqrt{u} G(u)$ and $u\geq u_r$ (by concavity of $H(u)$) and, in particular, $H(u)\leq 1 + \frac{3 u}{8}$ for $u\geq 0$; he showed how to obtain bounds on $g(r,x)$ in consequence. He also mentioned once a bound that is useful, if crude, viz., given by $G(u)\leq 1/\sqrt{u} + 1$ for $u\geq 0$.  (It can overestimate $G(u)$ by up to almost 50%, and $1+G(u)$ by up to almost 30%.) Let us work with this crude bound in detail.

We obtain
$$\begin{aligned}I(r,x)&\leq e^{-2 x}\int_{u_r}^\infty e^{-x u} \left(2 + \frac{1}{\sqrt{u}}\right) du = \frac{e^{-x (u_r+2)}}{x} + \frac{e^{-2 x}}{\sqrt{x}} \int_{x u_r}^\infty e^{-v} \frac{dv}{\sqrt{v}} \\ &= \frac{e^{-x (r+1/r)}}{x} + \frac{2 e^{-2 x}}{\sqrt{x}} \int_{\sqrt{xu_r}}^\infty e^{-t^2} dt
= \left(\frac{1}{x} + \frac{2}{\sqrt{x}} M(\sqrt{x u_r}) \right)e^{-x (r+1/r)},\end{aligned}$$
where $M(x)$ is Mills' ratio $M(x) = e^{x^2} \int_x^\infty e^{-t^2} dt$.

The special case $u_r=0$ in that, according to an answer by @openletter.mousetail.nl, it leads to a fancy special function called a leaky aquifer function (too fancy for the NIST handbook, apparently). Of course we can just use the bound above.


Just as above,
$$\begin{aligned}J(r,x) &= e^{-2 x} \int_0^{u_r} e^{-x u} \left(1 + \frac{1}{\sqrt{u}}\right) du = e^{-2 x} \frac{1-e^{-x u_r}}{x} + \frac{e^{-2 x}}{\sqrt{x}} \int_0^{x u_r} e^{-v} \frac{dv}{\sqrt{v}} \\ &= \frac{e^{-2 x}-e^{-x (r+1/r)}}{x} + \frac{2 e^{-2 x}}{\sqrt{x}} \int_0^{\sqrt{xu_r}} e^{-t^2} dt
\leq \frac{e^{-2 x}}{x} + \sqrt{\frac{\pi}{x}} e^{-2 x} -
\left(\frac{1}{x} + \frac{2}{\sqrt{x}} M(\sqrt{x u_r}) \right) e^{-x (r+1/r)}.\end{aligned}$$

We conclude that
$$g(r,x) = \begin{cases}\left(\frac{1}{2 x} + \frac{M(\sqrt{x u_r})}{\sqrt{x}}\right) e^{-x (r+1/r)} &\text{if $r\geq 1$,}\\ \left(\frac{1}{x} + \sqrt{\frac{\pi}{x}}\right) e^{-2 x} - \left(\frac{1}{2 x} + \frac{M(\sqrt{x u_r})}{\sqrt{x}}\right) e^{-x (r+1/r)}& \text{if $0<r<1$,}
\end{cases}$$

 For $x\geq 0$, $\frac{1}{x+\sqrt{x^2+2}} < M(x)\leq \frac{1}{x+\sqrt{x^2+(4/\pi)}}$ and $\frac{\sqrt{\pi}}{2 \sqrt{\pi} x + 2} \leq M(x)<\frac{1}{x+1}$ (NIST Handbook, section 7.8).

Of course we also have the upper bound $$g(r,x)\leq g(0,x) = \int_{0}^{\infty} e^{-x (s+1/s)} ds = 2 K_1(2 x),$$ where $K_1$ is the modified Bessel function of the second kind. The leading-order terms of $2 K_1(2 x)$ are $\sqrt{\frac{\pi}{x}} e^{-2 x} \left(1 + \frac{3}{8 x} + \dotsc\right)$. As @igorKhavkine points out, this is very good for $r$ close to $0$ (or $r\in (0,1)$ and $x$ large).

I'll add some remarks about an application in a bit.