What do you think of ?
This function is positive, continuous on , its abscissa of convergence is 0, and yet its Laplace transform at 0 is finite. Thus you don't have .
However, the first part is correct: 0 is a singularity for , which means that cannot be extended analytically to a neighbourhood of 0.
This is very very much reminiscent of the following result. Consider the power series , of radius . Assume for all . Then is a singularity of .
Actually, I noticed that the result about Laplace transforms can even be reduced to this one. Here is how. For simplicity (and wlog), assume . Then the function (mind the + sign) is analytic on the subset by definition of (and because of the change of sign) and the radius of convergence at 0 is exactly 1. Let us write inside the unit disk. Derivation in the integral gives , thus . We deduce (by the above mentioned result) that is a singularity for , hence is a singularity for .
As a conclusion, let me say a word about the (nice) proof of the result on power series. Take for simplicity. We procede by contradiction. Suppose that extends to a disk centered at 1. It is visibly obvious that the new domain (composed of the unit disk and this small disk around 1) contains a closed disk centered at and of radius . Thus is expandable in power series in this disk, hence if we choose , the series converges toward (where we're dealing with the extension of ). Now comes the nice part. The above series is and, because the coefficients are positive, we are allowed to change the order of summation, which gives (by Newton's binomial formula). Thus, we have proved that the series converges (toward the value ). This contradicts the fact that the radius of convergence is 1, since we could choose .