Take any real $a>e$. Then
\begin{equation*}
f(x)=f_{1,a}(x)+f_{2,a}(x), \tag{1}\label{1}
\end{equation*}
where
\begin{equation*}
f_{1,a}(x):=\int_e^a dt\,e^{-xg(t)},\quad
f_{2,a}(x):=\int_a^\infty dt\,e^{-xg(t)}, \tag{2}\label{2}
\end{equation*}
\begin{equation*}
g(t):=\frac t{\ln t}. \tag{3}\label{3}
\end{equation*}
Note that
\begin{equation*}
0\le f_{1,a}(x)\le a. \tag{4}\label{4}
\end{equation*}
The function $g\colon[e,\infty)\to[e,\infty)$ is continuous and increasing from $g(e)=e$ to $g(\infty-)=\infty$, with
\begin{equation*}
g'(t)=\frac{\ln t-1}{\ln^2 t}>0
\end{equation*}
for $t\in(e,\infty)$. So, $g$ has the inverse $h:=g^{-1}$, which is continuous and increasing from $h(e)=e$ to $h(\infty-)=\infty$, and also $h$ is differentiable on $(e,\infty)$. So, making the substitution $s=g(t)\iff t=h(s)$, we have
\begin{equation*}
f_{2,a}(x)=\int_{g(a)}^\infty ds\,h'(s)e^{-xs}=\int_{g(a)}^\infty ds\,\frac{\ln^2 h(s)}{\ln h(s)-1}e^{-xs}. \tag{5}\label{5}
\end{equation*}
Note that $h(s)\sim s\ln s$ as $s\to\infty$. So, there exists a function $A$ from a right-neighborhood of $0$ to $(e,\infty)$ such that
\begin{equation*}
A(x)\to\infty,\quad A(x)=o\Big(\frac1x\ln\frac1x\Big),\quad
xg(A(x))\to0, \tag{6}\label{6}
\end{equation*}
and
\begin{equation*}
\text{$h(s)\sim s\ln s$, and hence $\ln h(s)\sim\ln s$, } \\
\text{uniformly in $s\in[g(A(x)),\infty)$;}\tag{7}\label{7}
\end{equation*}
all limits here and in what follows are taken for $x\downarrow0$.
It follows by \eqref{5}, \eqref{6}, \eqref{7} that
\begin{equation*}
\begin{aligned}
f_{2,A(x)}(x)&\sim\int_{g(A(x))}^\infty ds\,\ln s\,e^{-xs} \\
&=\int_{xg(A(x))}^\infty \frac{du}x\,\Big(\ln u+\ln\frac1x\Big)\,e^{-u} \\
&=\frac1x\,\int_{xg(A(x))}^\infty du\,\ln u\,e^{-u}
+\frac1x\,\ln\frac1x\,\int_{xg(A(x))}^\infty du\,e^{-u} \\
&=\frac{1+o(1)}x\,\int_0^\infty du\,\ln u\,e^{-u}
+\frac{1+o(1)}x\,\ln\frac1x\,\int_0^\infty du\,e^{-u} \\
&\sim\frac1x\,\ln\frac1x.
\end{aligned}
\tag{8}\label{8}
\end{equation*}
It also follows from \eqref{4} and \eqref{6} that
\begin{equation}
f_{1,A(x)}(x)=o\Big(\frac1x\,\ln\frac1x\Big). \tag{9}\label{9}
\end{equation}
Finally, by \eqref{1}, \eqref{8}, \eqref{9},
\begin{equation}
f(x)\sim\frac1x\,\ln\frac1x.
\end{equation}