Don't solve it fully. Give me some hints to begin with. I tried one approach: Since we know $L[\int_{0}^{x} f(x) dx]=\frac{F(s)}{s}$ if $L[f(x)]= F(s)$. Thus I only need $L^{-1}[e^{1/s}]=g(x)$. And then $\int_{0}^{x} g(x) dx$ is the required inverse Laplace transform. But $g(x)$ does not exist because $\lim_{s\to\infty} e^{1/s} \neq 0$. So this doesn't work. I understand that I need to treat this $\frac{e^{1/s}}{s}$ as a whole. WolframAlpha returns $I_0(2\sqrt{x})$ as the answer which I can't even understand. What is that $I_0$? Please help me out.
EDIT: I am not allowed to use the expression for $L^{-1}f(x)$ as mentioned in one answer below. We aren't taught modified Bessel function also. What I can use: definition of Laplace transform, Laplace transform expressions for elementary functions, inverse Laplace transform by inspection (i.e. using the Laplace transform in reverse, basically observation), exponential shift, Convolution.
First of all, Wolfram Alpha tells you what $I_0$ is:
Secondly, note that by definition:
\begin{align} I_0(x) &= \frac{1}{2\pi i} \oint \frac{1}{t} e^{\frac{1}{2}x(\frac{1}{t}+t)} dt \\ \mathcal L^{-1}f(x) &=\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} \frac{1}{s} e^{\frac{1}{s} + sx} ds\end{align}
Observe what happens to the second integral after the transformation $s = \frac{t}{\sqrt{x}} $. After that think about why the line integral must be equal to the path integral around the origin.