(Linear, homogeneous, constant-coefficient) differential equations are already polynomial equations. That is, let $D$ denote the differential operator $\frac{d}{dx}$. Then linear homogeneous constant-coefficient differential equations are nothing more than equations of the form $p(D) y = 0$, where $y$ is some function of $x$ and $p$ is some polynomial. Writing $p(x) = \prod_{i=1}^n (x - r_i)^{e_i}$ where the roots $r_i$ are distinct, it follows that we can write the equation as
$(D - r_1)^{e_1} (D - r_2)^{e_2} ... (D - r_n)^{e_n} y = 0.$
The point here is that $D$ is a linear operator, and it generates a commutative algebra of differential operators. Now, from the above it follows that any solution to $(D - r_i)^{e_i} y = 0$ is a solution to the above, and it turns out that linear combinations of these give all solutions. (This is a general fact about linear operators on finite-dimensional vector spaces; see, for example, Axler's Linear Algebra Done Right.)
In the special case that $e_i = 1$ for all $i$ things are particularly simple: $(D - r_i) y = 0$ just means $Dy = r_i y$, or $y = c_i e^{r_i x}$ for some constant $c_i$. This gives $n$ linearly independent solutions, which is the correct number, hence the general solution has the form
$y = \sum_{i=1}^n c_i e^{r_i x}.$
If the multiplicities $e_i$ are higher then we must consider solutions to $(D - r_i)^{e_i} y = 0$. Let $w(x) = e^{-r_i x} y(x)$. Then $D w = e^{-r_i x} (D - r_i) y$, from which it follows that the above holds if and only if $D^{e_i} w = 0$. But this says precisely that $w$ is a polynomial of degree at most $e_i - 1$. Hence
$y = c_i(x) e^{r_i x}$
where $c_i(x)$ is a polynomial of degree at most $e_i - 1$. This gives $e_i$ linearly independent solutions, so again taking linear combinations we have found all of them.
None of this requires that you know anything about integral transforms. From the abstract point of view, integral transforms are just a convenient way to diagonalize $D$. The Laplace transform (more or less) takes $D$ to multiplication by $s$, hence turns $p(D)$ into $p(s)$ (the roots of which are the eigenvalues $r_i$ above), but this is just a more concrete way of realizing the abstract structure already evident in the problem: that the set of solutions to $p(D) y = 0$ is naturally acted on by $\mathbb{C}[D]$.