in general, assuming that the solution has a Laplace transform, aren't we restricting ourselves to a subset of the solution set of the ODE? My professor in a lecture on ODEs, stated the following:

Consider the following ODE $$y'' - 2y = f(x),$$ where the Laplace
  transform of $f(x)$ exists.
To solve this ODE, we can argue that; Assume the solution of this ODE
  has a Laplace transform and by using the properties of LP,  $$L(y''
 -2y) = L(f(x))$$ [...] hence we get an algebraic equation instead of an DE.

I mean as far as I know this is a classical argument that is made while solving an ODE with the method of Laplace (or any other) transform method.
However, in general, assuming that the solution has a Laplace transform, aren't we restricting ourselves to a subset of the solution set of the ODE ?
I mean, in general, can't the ODE have a solution that does not have a Laplace transform ?
 A: A priori yes, we could be missing solutions this way, but it's not hard to prove that it's actually ok.
Say $E$ is the space of functions defined on $(0,\infty)$ that "have Laplace transforms", meaning that there exist $c$ and $a$ so $|f(t)|\le ce^{at}$. For the specific DE you mention we need to show that if $y''-2y\in E$ then $y\in E$.
First note that 


Lemma If $b$ is a complex constant and $y'-by\in E$ then $y\in E$. 


This follows from the standard algorithm for solving $y'-by=f$ using integrating factors.
Now if $D$ denotes differentiation and $I$ is the identity operator then $$y''-2y=(D^2-2I)y.$$And $$D^2-2I=(D+\sqrt 2I)(D-\sqrt 2I).$$
Now assume that $y''-2y\in E$. This says that $$(D+\sqrt 2I)(D-\sqrt 2I)y\in E.$$The Lemma shows that $$(D-\sqrt  2I)y\in E,$$and now a second application of the Lemma shows that $y\in E$.
Similarly for any constant-coefficient linear DE, by the fundamental theorem of algebra.
Edit: Another version of the samme argument, for readers who are confused by that $D^2-2I=(D+\sqrt 2I)(D-\sqrt 2I)$ thing:
Assume $f\in E$ and $y''-2y=f$. Define $$z=y'-\sqrt 2y.$$Simply working out the derivative you can check that $$z'+\sqrt 2z=y''-2y=f.$$So the Lemma shows that $z\in E$, and now since $y'-\sqrt 2y=z$ another application of the Lemma shows that $y\in E$.
Q: Why didn't I just say that in the first place?A: It's more clear how the first version generalizes to other DEs, by just factoring the relevant polynomial. Abstraction is useful...
