My language of mathematics can be sloppy as I have a physics background.
The Fourier transformable functions $f(x)$ (which obeys absolute integrability condition $\int\limits_{-\infty}^{\infty} |f(x)|dx<\infty$) can be decomposed as $$f(x)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}\tilde{f}(k)e^{ikx}dk\tag{1}$$ which implies that for such functions $f(x)$, the set $\{e^{ikx}\}$ where $k$ varies continuously over $-\infty$ to $+\infty$ forms a complete set. Similarly functions $g(x)$ which are Laplace transformable can be written as $$g(t)=\int\limits_{0}^{\infty}\tilde{g}(s)e^{-st}ds.\tag{2}$$
Does it not mean that the set $\{e^{-st}\}$ where $s$ varies continuously over $0$ to $+\infty$ also form a complete set for the functions $g(x)$?
The OP has said that (s)he is going to delete the question - since that hasn't happened yet, there may as well be an answer for the record:
None of this makes much sense, as far as I can see: The question asks something about the Laplace transform, by analogy with the Fourier transform, but the previous facts about the Fourier transform are simply not so.
First, assuming just $\int|f(t)|\,dt<\infty$ does not imply that $f$ has a representation $$f(x)=\frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^{\infty}\tilde{f}(k)e^{ikx}dk.$$That would be true if $\int|f|<\infty$ and $\int|\hat f|<\infty$.
Does this imply that the functions $e^{ikt}$ form a "complete set"? The OP clarified in a comment that here "complete set" means "complete set in $L^2$". The functions $e^{ikt}$ are certainly not a complete set in $L^2$, since they're obviously not even elements of $L^2$ to begin with.
Now we come to the question about "Laplace transformable" functions. My best guess, neither confirmed nor denied by the OP, is that saying $g$ is Laplace tranformable means that $g$ has a Laplace transform. If so then saying $g$ is Laplace transformable does not say that $$g(s)=\int_0^\infty\tilde g(t)e^{-st}\,dt;$$(or if it does this is news to me) - that says that $g$ is a Laplace tranform (in particular it is the Laplace transform of $\tilde g$), not that $g$ has a Laplace transform.
(No, come to think of it saying $g$ has a Laplace transform clearly does not imply $g$ is a Laplace transform. For example, under minimal growth conditions in $h$, if $g=\mathcal L\{h\}$ then $g$ is continuous.)