Consider the Laplace transform of $e^x$, given by
$$ \int_0^\infty e^x e^{-sx} dx = \int_0^\infty e^{x(1-s)} dx = \int_0^\infty e^{-x(s-1)} dx$$
When written like the integral in the middle, the integral appears to diverge whereas the integral on the right clearly converges, but this is the exact same integral!
Why does this "hidden convergence" occur? If you were looking at this integral, you might not see something so obvious as to pull the negative out and voila. How would I know that this integral must converge even though it appears to diverge before pulling out the negative?
You are confused because you consider $\int_0^\infty e^x e^{-sx} dx $ as an unique integral, which is not the case. In fact, this it is a family of integrals : For each value of $s$ there is an integral different from the others.
As such it is not surprising that this family of integrals includes some which are convergent and some which are not convergent.
The real question is to determine for which values of $s$ the corresponding integrals are convergent, while for the other values of $s$ the other integrals are not convergent. I suppose that it is easy for you to answer to this question.