I've been thinking about this since my original comment actually, but there are complexities from using the noncompact group R. The exponential functions are still the simultaneous eigenvectors of time shifts.
The Laplace transform seems to be just using the fact that <f,g> = integrate(f(x) g(x), x from 0 to infinity) is an inner product for the space of square integrable functions (probably better would be <f,g> = integrate(f(x) conj(g(x)), x from 0 to infinity) as a Hermitian product). The various e^(ax) functions are linearly independent, so the functions g |-> <e^(ax), g> are linearly independent functionals. If the exponential functions are actually enough, then this means you can study a function by studying the vector consisting of its value through all the functionals, which is the Laplace transform.
The Laplace transform has a pretty bad inverse formula, partly because the exponential functions are not orthogonal with respect to the inner product.
The Laplace transform seems to be just using the fact that <f,g> = integrate(f(x) g(x), x from 0 to infinity) is an inner product for the space of square integrable functions (probably better would be <f,g> = integrate(f(x) conj(g(x)), x from 0 to infinity) as a Hermitian product). The various e^(ax) functions are linearly independent, so the functions g |-> <e^(ax), g> are linearly independent functionals. If the exponential functions are actually enough, then this means you can study a function by studying the vector consisting of its value through all the functionals, which is the Laplace transform.
The Laplace transform has a pretty bad inverse formula, partly because the exponential functions are not orthogonal with respect to the inner product.