Can you control the oscillation of f(x) as x increases? If you can show that the ratio of f(x) to your 'simplified' form is 'slowly varying' then your asymptotics will probably work out.
A simple example of what you cannot afford is a log-periodic oscillation; this is because the limits of oscillation of the function and Laplace transform need not agree. The simplest example of a log-periodic oscillation is a complex exponential:
$\int_{0}^{\infty }e^{-st}t^{\alpha +i\beta }dt=\left[ \frac{\Gamma \left( \alpha +i\beta +1\right) }{\Gamma \left( \alpha +1\right) }e^{i\beta \log t}\right] \frac{\Gamma \left( \alpha +1\right) }{s}s^{-\alpha }$
In a sense you can view the imaginary part of the exponential as a 'wobbly constant' which changes more and more slowly. The point is the amplitude of the wobble in the transform depends on β but not for the function.
If you do have this sort of problem (it happens all the time in analysis of algorithms and chaotic dynamics) then you can for example resort to the 'gamma function method' of DeBruijn.
The same thing holds true for the moment question. If you look up a counterexample for the moment problem, (e.g. Feller volume II p. 227) you see the ubiquitous log-periodic oscillation.
Not surprisingly the log-periodic oscillation also shows up in convergence questions of Fourier series, but there it is not oscillating more and more slowly, but faster and faster.
As long as the singularities are not "too bad", the answer will be yes, $f(x)$ will represent a solution of the differential equation. The very same way that
$$ \sum_{n=0}^{\infty} (-1)^nn!x^{n+1} $$
represents a solution of $x^2y'+y=x$. One might object that the series diverges, but resummation theory says this is irrelevant, that sum nevertheless represents a unique function (in a sector) which is a solution of the differential equation.
Balser's book, "From divergent series to analytic differential equations", would be a good place to start. Since the theory fundamentally uses Mellin transforms, you should be able to find what you want (perhaps indirectly) there.
Best Answer
What is also very interesting is that the Laplace transform is nothing else but the continous version of power series - see this insighful video lecture from MIT:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-19-introduction-to-the-laplace-transform/