$\forall\ x\ \in\ \left]0,+\infty\right[\ $ we put: $$ {\rm f}\left(x\right) = \int_{0}^{\infty}{{\rm e}^{-xt} \over \left(1 + t^{3}\right)^{1/3}}\,{\rm d}t $$ The question is the question is to find an equivalent of $\,\,{\rm f}\left(x\right)$ when $x \to 0^{+}$.
That means find a simple function $g$ such that when $x \to 0^+$ we have : $f(x)\sim g(x)$ that means : $$\lim_{x \to 0^+} \frac{f(x)}{g(x)} =1$$
$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{\rm f}\left(x\right) = \int_{0}^{\infty}{{\rm e}^{-xt} \over \left(1 + t^{3}\right)^{1/3}}\,{\rm d}t}$
When $x \gg 1$: \begin{align} {\rm f}\left(x\right) &= \int_{0}^{\infty} {{\rm e}^{-t} \over x\left[1 + \pars{t/x}^{3}\right]^{1/3}}\,{\rm d}t = {1 \over x}\int_{0}^{\infty}\expo{-t} \sum_{n = 0}^{\infty}{1/3 \choose n}\bracks{\pars{t \over x}^{3}}^{n}\,\dd t \\[3mm]&= {1 \over x}\sum_{n = 0}^{\infty}{1/3 \choose n}{1 \over x^{3n}}\int_{0}^{\infty} \expo{-t}t^{3n}\,\dd t = {1 \over x}\sum_{n = 0}^{\infty}{1/3 \choose n}{\pars{3n}! \over x^{3n}} = {1 \over x}\sum_{n = 0}^{\infty} {\Gamma\pars{4/3}\pars{3n}! \over n!\Gamma\pars{4/3 - n}}\,{1 \over x^{3n}} \\[3mm]&= {1 \over x}\pars{1 + {2 \over x^{3}} + \cdots} \end{align}
$$ \mbox{For example, we can take}\ \color{#0000ff}{\large{\rm g}\pars{x} = {1 \over x}} $$
Mathematica gives $$ \frac{G_{1,4}^{4,1}\left(\frac{s^3}{27}| \begin{array}{c} \frac{2}{3} \\ 0,0,\frac{1}{3},\frac{2}{3} \\ \end{array} \right)}{2 \sqrt{3} \pi \Gamma \left(\frac{1}{3}\right)}$$ for the integral, and expanding this in a power series at $0,$ gets $$ \frac{-3 \sqrt{3} \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{2}{3}\right) \log (s)-2 \sqrt{3} \gamma \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{2}{3}\right)+\sqrt{3} \log (27) \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{2}{3}\right)+\sqrt{3} \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{2}{3}\right) \psi ^{(0)}\left(\frac{2}{3}\right)}{6 \pi }+\frac{4 \pi ^2 s}{27 \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{4}{3}\right)^2}+O\left(s^2\right), $$ so the dominant term is $$ \frac{-3 \sqrt{3} \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{2}{3}\right)}{6\pi} \log (s) $$
Running FullSimplify on the power series, gets one $$ \left(-\log (s)+\frac{\pi }{6 \sqrt{3}}-\gamma +\frac{\log (3)}{2}\right)+\frac{4 \pi ^2 s}{27 \Gamma \left(\frac{1}{3}\right) \Gamma \left(\frac{4}{3}\right)^2}+O\left(s^2\right), $$ which agrees with Robert Israel's comment.
Set $$ h(x,t)=\left\{ \begin{array}{lll} \mathrm{e}^{-tx} & \text{if} & t\in[0,1), \\ \frac{\mathrm{e}^{-tx}}{t} & \text{if} & t\in [1,\infty), \end{array} \right. $$ and $g(x)=\int_0^\infty h(x,t)\,dt$.
