Does $L^2([0,\infty))$ have a Hilbert basis consisting of "nice" exponentials, $a_ne^{b_nx}, a_n>0, b_n\in\mathbb{C}$, where "nice" means that $a_n$ and $b_n$ are given by a simple, possibly recursive, formula?
Do the exponentials $\{e^{s x}\}_{s\in\mathbb{C}}$ even span $L^2([0,\infty))$? I feel like I'm missing something here.
For a discrete decomposition, the closest natural/well-studied thing I can think of would be functions of the form $H_{2n}(x)\,e^{-x^2/2}$, where $H_n$ are Hermite polynomials, and we only need the even-index ones. Up to normalizations these give an orthonormal basis of $L^2[0,+\infty)$. Similarly, one could use $H_{2n+1}(x)e^{-x^2/2}$. Differentiation maps between the two orthogonal families.
These are eigenfunctions for the "quantum harmonic oscillator" $-{\partial^2\over \partial x^2}+x^2$, which has compact resolvent, and is essentially self-adjoint, so gives $L^2(\mathbb R)$ an orthogonal basis of eigenfunctions. The so-called (google-able) "ladder operators" $i{\partial\over \partial x}\pm ix$ exactly map (up to determinable constants) $H_ne^{-x^2/2}\to H_{n\mp 1}e^{-x^2/2}$. (Yes, note the opposite-ness of signs.)