The bilateral Laplace transform is defined as,
$$X(s) = \int_{-\infty}^{\infty}x(t)e^{-st}dt~~,s = \sigma +j\omega$$
where both $\sigma$ and $\omega$ are real. Then,
$$X(s) = X(\sigma+j\omega) = \int_{-\infty}^{\infty}x(t)e^{-\sigma t}e^{-j\omega t}dt \leq\int_{-\infty}^{\infty}|x(t)e^{-\sigma t}e^{-j\omega t}|dt = \int_{-\infty}^{\infty}|x(t)|e^{-\sigma t}dt$$
So we can deduce that,
$$\int_{-\infty}^{\infty}|x(t)|e^{-\sigma t}dt < \infty \Rightarrow X(s)~~~\text{converges if}~~ \Re\{s\} = \sigma$$
What about the reverse? That is, if $X(s)$ is convergent then $x(t)\exp(-\Re\{s\}t)$ is absolutely integrable.
Can you please provide either a proof or a counter example. The book Signals and Systems by Oppenheim and Wilsky uses that as a fact but I am not convinced.