I know of no such publication. However, this issue may simply be on one hand too trivial and on the other hand too far removed from practical relevance. Let's derive what you are after to see:
A creation ladder operator $\hat{a}^\dagger$ for arbitrary states would have to be of the form $$\sum_{n=0}^\infty c_n \left| n+1 \right\rangle \left\langle n \right|$$ where you might allow a freedom to choose coefficients $c_n$ to try fulfilling whatever useful features of $\hat{a}^\dagger$ and $\hat{a}$ you may desire, say some generalization of the number operator $\hat{a}^\dagger \hat{a}$. You might choose to have its eigenvalues be the quantum number $n$ or the energy divided by the energy difference between ground state and first excited state—either, and probably some other choices are sensible generalizations of the situation for the harmonic oscillator. In fact, by choosing $c_n = \sqrt{n+1}$ and a harmonic oscillator Hamiltonian or, then, $\hat{H}=\hbar \, \omega \, \big( \hat{a}^\dagger \hat{a} + \frac{1}{2} \big)$, you reproduce the equations you gave in the question in a simpler notation abstracting away from position and momentum representation.
In general, $\hat{a}^\dagger$ (possibly through the $c_n$ and certainly through the energy eigenstates $\left| n \right\rangle$) will depend not just on your choice of what you want from the number operator, but crucially also on your system's Hamiltonian. The straight-forward way to derive $\hat{a}^\dagger$ is to solve the Hamiltonian for the energy eigenstates—which is the way you would have to go to get ladder operators from first principles if it were not for textbooks simply dropping a definition on you, (possibly) like the one you reproduced. You would simply insert the energy eigenstates $\left| n \right\rangle$ and formulate the constraint for the $c_n$ and solve for it. The simplest example, because the $c_n$ then do not depend on the Hamiltonian, is the constraint that the number operator has the quantum number as eigenvalues: $$\left\langle n \right| \hat{a}^\dagger \hat{a} \left| n \right\rangle = n \qquad \Rightarrow \qquad c_n = \sqrt{n+1}$$
That means that the ladder operators for any specific (non-harmonic oscillator) Hamiltonian is usually not helpful in solving this Hamiltonian but at best for re-expressing it. A harmonic oscillator is a rare case where this leads to a simplification, but this will not be the case in general. The fact that you obviously cannot retain all useful features e.g. of the number operator further limits the utility of specialized ladder operators.
