= Solution
The expansion is singular near the moving lower limit, so expanding $e^{-\varepsilon/x}$ throughout the original <integral> would lose its constant term. Introduce the symmetric <exponential integral>
$$
J(\varepsilon)=\int_0^\infty\frac{e^{-x-\varepsilon/x}}x\,dx.
$$
Under $x=\varepsilon/t$, the discarded piece becomes
$$
J-I=\int_1^\infty\frac{e^{-t-\varepsilon/t}}t\,dt=\beta-\varepsilon\int_1^\infty\frac{e^{-t}}{t^2}\,dt+O(\varepsilon^2)=\beta-\varepsilon(e^{-1}-\beta)+O(\varepsilon^2).
$$
This is a uniform <Taylor expansion> because $t\geq1$. To derive the <logarithmic expansion of a symmetric exponential integral> without needing another special function, first exploit the same substitution to write
$$
J=2\int_{\sqrt\varepsilon}^\infty\frac{e^{-x-\varepsilon/x}}x\,dx=2E(\sqrt\varepsilon)+O(\sqrt\varepsilon)=-\log\varepsilon-2\gamma+o(1).
$$
The estimate follows from $0\leq1-e^{-\varepsilon/x}\leq\varepsilon/x$. Here $\gamma$ is <Euler's constant> and the <small-argument expansion of the exponential integral> fixes the constant as well as the logarithm.
Differentiate under the integral and integrate the derivative of $e^{-x-\varepsilon/x}/x$ over $(0,\infty)$; the exponentially small endpoint terms vanish. This gives the <ordinary differential equation>
$$
\varepsilon J''+J'=J.
$$
For the <Frobenius method> at its <regular singular point>, write $J=A\log\varepsilon+B+\varepsilon(C_1\log\varepsilon+D)+\cdots$. Substitution gives $C_1=A$ and $D=B-2A$. The preceding leading calculation sets $A=-1$, $B=-2\gamma$, hence
$$
J=-\log\varepsilon-2\gamma+\varepsilon[-\log\varepsilon+2-2\gamma]+O(\varepsilon^2|\log\varepsilon|).
$$
Subtracting the <moving-cutoff correction to a symmetric exponential integral> yields the requested answer, including the logarithmically enhanced first-order term:
$$
\boxed{I(\varepsilon)=-\log\varepsilon-2\gamma-\beta+\varepsilon[-\log\varepsilon+2-2\gamma+e^{-1}-\beta]+O(\varepsilon^2|\log\varepsilon|).}
$$
Thus the $\varepsilon\log\varepsilon$ <switchback term> must be retained when asking for all contributions through order $\varepsilon$.
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