Well, there are two contributions because the exponent is zero at $t=0$ and $t=\pi/2$. Let's consider $t=0$ first. In the immediate neighborhood of $t=0$ (we'll get to what that means in a bit), the exponent behaves as $x t^3$ so that as $x\to\infty$, we have
$$I(x) \sim \int_0^{\infty} dt \, e^{-x t^3} = \frac{\Gamma\left ( \frac{4}{3}\right )}{x^{1/3}} = \frac{\Gamma\left ( \frac{1}{3}\right )}{3 x^{1/3}} \quad (x\to\infty)$$
What is the size of the neighborhood? Well, we want $0 \lt x t^3 \lt \epsilon$ for some small $\epsilon$, so we have $0 \lt t \lt (\epsilon/x)^{1/3}$.
We also have a contribution in a neighborhood near $t=\pi/2$; we Taylor expand and get that
$$t^3 \cos{t} = -\frac{\pi^3}{8} \left ( t-\frac{\pi}{2}\right ) + O\left [ \left ( t-\frac{\pi}{2}\right )^2\right ]$$
Really, we are only interested in positive values of this expansion, as the exponent is positive through the integration region. If we look at only an immediate neighborhood near $t=\pi/2$, but with $t \lt \pi/2$, then we may approximate the contribution to the integral there as
$$\int_0^{\infty} dy \, e^{-\pi^3 x y/8} = \frac{8}{\pi^3 x}$$
It doesn't make sense to simply add these two terms together and declare them the leading behavior of $I(x)$ until we investigate the next leading behavior of the contribution at $x=0$. Note that the next contribution in the exponent is $x t^5/2$; within the interval of interest, this is $O(x^{-2/3})$, so we may Taylor expand this exponential term separately. The result is the following integral for the next contribution at $t=0$:
$$\frac12 x\int_0^{\infty} dt \;t^5 \, e^{-x t^3} = \frac1{6 x}$$
Note that this is $O(1/x)$ as is the leading contribution from $t=\pi/2$, so we may add these. The stated result follows.
Best Answer
Write $\varphi(x) = \exp(1/\log x)$ and note that $\varphi^{-1} = \varphi$. So by substituting $y = \varphi(t)$, or equivalently $t = \varphi(y)$,
\begin{align*} I(x) := \int_{0}^{x} \exp\left(\frac{1}{\log t}\right) \, \mathrm{d}y &= \int_{1}^{\varphi(x)} y \varphi'(y) \, \mathrm{d}y \\ &= \left[ y \varphi(y) \right]_{1}^{\varphi(x)} + \int_{\varphi(x)}^{1} \varphi(y) \, \mathrm{d}y \\ &= c + x \varphi(x) - \int_{0}^{\varphi(x)} \varphi(y) \, \mathrm{d}y, \end{align*}
where
$$c := \int_{0}^{1} \varphi(y) \, \mathrm{d}y \approx 0.27973176.$$
Using the expansion $ \varphi(y) = \sum_{n=0}^{\infty} \frac{1}{n!(\log y)^n} $ which converges uniformly for $0 < y \leq \varphi(x)$, we can perform term-wise integration, obtaining
\begin{align*} I(x) &= c + x \varphi(x) - \sum_{n=0}^{\infty} \int_{0}^{\varphi(x)} \frac{1}{n!(\log y)^n} \, \mathrm{d}y \\ &= c - (1-x) \varphi(x) - \sum_{n=0}^{\infty} \frac{1}{(n+1)!} \int_{0}^{\varphi(x)} \frac{\mathrm{d}y}{(\log y)^{n+1}} \end{align*}
Now by using the integral formula
$$ \int \frac{\mathrm{d}y}{(\log y)^{n+1}} = \frac{1}{n!} \left( \operatorname{li}(y) - \sum_{k=1}^{n} \frac{(k-1)! y}{(\log y)^k} \right), $$
where $\operatorname{li}(x) = \int_{0}^{x} \frac{\mathrm{d}t}{\log t}$ is the logarithmic integral function, we get
\begin{align*} I(x) &= c - (1-x) \varphi(x) - \sum_{n=0}^{\infty} \frac{1}{n!(n+1)!} \left( \operatorname{li}(\varphi(x)) - \varphi(x) \sum_{k=1}^{n} (k-1)! (\log x)^k \right) \\ &= \boxed{ c - a_0 \operatorname{li}(\varphi(x)) + \varphi(x) \sum_{k=1}^{\infty} (k-1)! a_k (\log x)^k }, \tag{*} \end{align*}
where
$$ a_k := \sum_{n=0}^{\infty} \frac{1}{(n+k)!(n+k+1)!}. $$