"Proving" the definition of the Laplace Transform from two properties Suppose $\mathbf{G}$ is the class of piecewise continuous functions of exponential order from $[0,\infty)$ to $\mathbb{R}$ and that $\mathcal{L}$ is a transformation defined on elements of $\mathbf{G}$ satisfying the two properties  

* For $f,\,g\in\mathbf{G}$ and $c_1,\,c_2\in\mathbb{R}, \mathcal{L}\{c_1f+c_2g\}=c_1\mathcal{L}\{f\}+c_2\mathcal{L}\{g\}$ 

* $\mathcal{L}\{f^\prime(t)\}=s\mathcal{L}\{f(t)\}-f(0)$
Certain elementary properties of Laplace transforms follow easily from these two.
For example:
$\mathcal{L}\{0\}$ follows from property ($1$).
Let $f(t)=1$ for $t\ge0$. Then $\mathcal{L}\{1^\prime\}=s\mathcal{L}\{1\}-1$ therefore
$$\mathcal{L}\{1\}=\dfrac{1}{s}\tag{1}$$
Using this as a basis step and using property ($2$) to prove that for $n\ge0$
$$ \mathcal{L}\{t^{n+1}\}=\dfrac{s}{n+1}\mathcal{L}\{t^n\}$$
one may establish the inductive step showing that
$$\mathcal{L}\{t^n\}=\dfrac{n!}{s^{n+1}}\tag{2}$$
Also, since property ($2$) gives $\mathcal{L}\{\left(e^{at}\right)^\prime\}=s\mathcal{L}\{e^{at}\}-e^0$ we immediately get
$$ \mathcal{L}\{e^{at}\}=\dfrac{1}{s-a}\tag{3}$$
From $\mathcal{L}\{(\cos t)^\prime\}=s\mathcal{L}\{\cos t\}-1$ we obtain
$$ \mathcal{L}\{\sin t\}+s\mathcal{L}\{\cos t\}=1$$
and from $\mathcal{L}\{(\sin t)^\prime\}=s\mathcal{L}\{\sin t\}-0$ we obtain
$$ s\mathcal{L}\{\sin t\}-\mathcal{L}\{\cos t\}=0$$
And the solution of these two gives
$$ \mathcal{L}\{\sin at\}=\dfrac{a}{s^2+a^2}\tag{4}$$
$$ \mathcal{L}\{\cos at\}=\dfrac{s}{s^2+a^2}\tag{5}$$
When I first adopted this approach decades ago in a differential equations class I believed I also had a proof that 
$$ \mathcal{L}\{f(t)\}=\int_0^\infty e^{-st}f(t)\,dt$$
but I cannot recall it so I may have been mistaken.
Can anyone prove this usual definition of the Laplace transform from the two properties at the top of this post?
 A: Here's a Heaviside-like heuristic proof, which makes some assumptions about smoothness and continuity (it's probably possible to tighten this up into a genuine proof, but this is a first approximation).
Suppose $f$ is an analytic function in $\mathbf{G}$, with a power series that converges everywhere, and $\mathcal{L}[f](s)$ finite in a neighbourhood of $s=0$ (there are plenty of these, such as sufficiently negative exponentials). Then, by expanding in a power series about the origin, we find
$$ \mathcal{L}[xf](s) = -\mathcal{L}[f]'(s), $$
and further, induction gives
$$ \mathcal{L}\left[ x^n f\right](s) = \left(-D\right)^n \mathcal{L}[f](s), $$
where $D=d/ds$.
Now, the important thing to know next is that for smooth functions,
$$ e^{hD} F(s) = \sum_{n=0}^{\infty} \frac{h^n }{n!} D^n F(s) = \sum_{n=0}^{\infty} \frac{h^n }{n!} F^{(n)}(s) = f(s+h). $$
Then by linearity,
$$ \mathcal{L}\left[ e^{-ax} f\right](s) = \mathcal{L}\left[ \sum_{n=0}^{\infty} (-a)^n\frac{x^n}{n!} f\right](s) = \sum_{n=0}^{\infty} \frac{(-a)^n}{n!} \mathcal{L}\left[x^n f\right](s) = \sum_{n=0}^{\infty} \frac{(-a)^n}{n!}(-D)^n \mathcal{L}[f](s) = e^{aD} \mathcal{L}[f](s) = \mathcal{L}[f](s+a) $$
It remains to determine $\mathcal{L}[f]$ at one point; the way we set things up, this may as well be $s=0$. The derivative property gives, if $\mathcal{L}[f](0)$ and $\mathcal{L}[f'](0)$ exist,
$$ \mathcal{L}[f'](0) = -f(0). $$
There is, of course, another well-known linear operator that does this: $\int_0^{\infty} f'(x) \, dx = -f(0)$. It follows that where both operators are defined, they are equal, viz.
$$ \mathcal{L}[f](s) = \int_0^{\infty} e^{-sx} f(x) \, dx, $$
although again, one needs to be careful about convergence and so on.
