Harmonic oscillator
For the harmonic oscillator ladder operators the reason is the following:
A physical system where there is no lower bound on the energy is unstable.${}^{1}$ This is why in QM we always assume that there is one state which attains the smallest values of the energy: $|\Omega\rangle$.
As a consequence of the commutation relations, the energy of $\hat{a}|\Omega\rangle$ is one unit of $\omega \hbar$ less than that of $|\Omega\rangle$.
$$
H \,\hat{a}|\Omega\rangle = (E_{\mathrm{min}} -\hbar\omega)\,\hat{a}|\Omega\rangle\,.
$$
So the only way out is that $\hat{a}|\Omega\rangle$ is not an eigenstate, therefore it has to be the zero state (not $0$ as a number but the null element of the Hilbert space).
Angular momentum
For the angular momentum the reason is the following:
We want to build finite dimensional representations. For the same reasoning as above the eigenvalue under $\hat{L}_z$ of $\hat{L}_-|\Omega\rangle$ is $\hbar$ less than that of $|\Omega\rangle$. Since we are diagonalizing an Hermitian operator ($\hat{L}_z$), then $\hat{L}_-|\Omega\rangle$ is orthogonal to $|\Omega\rangle$, so it's either a new state or the zero state.
If it's not the zero state then this procedure never ends and the representation becomes infinite dimensional. There is nothing wrong with infinite dimensional representations (they do exist), but particles typically transform under finite dimensional ones.
Note: the two arguments are really the same argument, but I wanted to emphasize the fact that the crucial point (boundedness of energies vs. finite dimension) is different.
$\quad{}^1$ Having negative energies is not problematic per se. The theory is fine as long as there is a lower bound. Naturally we can always change the offset so that this lower bound is zero