As stated in wikipedia:
$$\mathcal{L}\{f(t)\}=\int_{0^-}^{\infty}e^{-st}f(t)dt$$ $$=\left[\frac{f(t)e^{-st}}{-s}\right]_{0^-}^{\infty}-\int_{0^-}^{\infty}\frac{e^{-st}}{-s}f'(t)dt\space,\space\space\space(by\space parts)$$ $$=\left[-\frac{f(0^-)}{-s}\right]+\frac{1}{s}\mathcal{L}\{f'(t)\}$$
Note that from the second line to the third, it's assumed that $t\to\infty$ makes the first term go to zero. Shouldn't this be specified only if $Re(s)>0$ and if $f(t)$ grows less than $e^{-st}$?
You should check again, but if, for instance, $f$ is a probability density function then the Laplace transform is well defined only if $e^{-st}f(t) \to 0$ for $t \to \infty$ which gives you line 3.