= Solution
Factor out the elementary exponential and write
$$
I(\lambda)=\lambda^{-1/2}e^{2\lambda}J(\lambda),
\qquad
J(\lambda)=\int_0^\infty
e^{-s-\lambda^2/s}\,ds.
$$
Use a <matched asymptotic expansion> with $0<\lambda\ll\delta\ll1$. In the outer part,
$$
J_{\rm out}
=\int_{\delta^2}^\infty e^{-s}
\left(1-\frac{\lambda^2}{s}+\cdots\right)ds.
$$
The given identity for the <Euler--Mascheroni constant>, integrated by parts, implies
$$
\int_{\delta^2}^\infty\frac{e^{-s}}s\,ds
=-\gamma_e-2\log\delta+O(\delta^2).
$$
Hence
$$
J_{\rm out}
=1-\delta^2
+\lambda^2(\gamma_e+2\log\delta)
+o(\lambda^2),
$$
apart from pure higher matching powers in $\delta$.
In the inner part, use $t=\lambda^2/s$:
$$
J_{\rm in}
=\lambda^2
\int_{\lambda^2/\delta^2}^\infty
t^{-2}e^{-t}e^{-\lambda^2/t}\,dt.
$$
Since $s=\lambda^2/t\leq\delta^2$, expand the last exponential. At the required order,
$$
\int_a^\infty t^{-2}e^{-t}\,dt
=\frac{e^{-a}}a-int_a^\infty\frac{e^{-t}}t\,dt
=\frac1a-1+\gamma_e+\log a+O(a).
$$
With $a=\lambda^2/\delta^2$, this gives
$$
J_{\rm in}
=\delta^2
+\lambda^2
\left[-1+\gamma_e
+2\log\lambda-2\log\delta\right]
+o(\lambda^2).
$$
The overlap terms $\delta^2$ and $\log\delta$ cancel, leaving the logarithmic <switchback term>
$$
J(\lambda)
=1+\lambda^2
\left(2\log\lambda+2\gamma_e-1\right)
+o(\lambda^2).
$$
Finally,
$$
e^{2\lambda}=1+2\lambda+2\lambda^2+O(\lambda^3),
$$
so
$$
\boxed{
I(\lambda)
\sim\lambda^{-1/2}
+2\lambda^{1/2}
+\lambda^{3/2}
\left(1+2\log\lambda+2\gamma_e\right)}.
$$
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