The ladder operators are not axiomatic. They are derived by a quite long chain of conclusions from the commutation relations. $$\begin{align} [L_x,L_y]=i\hbar L_z \\ [L_y,L_z]=i\hbar L_x \\ [L_z,L_x]=i\hbar L_y \end{align} \tag{1}$$ For angular momentum $\mathbf{L}$ these relations (1) can be derived from the definition $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ and the canonic commutators $[r_i,p_j]=i\hbar\delta_{ij}$. But for spin angular momentum $\mathbf{S}$ these relations cannot be derived from anything. So for spin we need to postulate (1) as an axiom. And then we can show that the total angular momentum $\mathbf{J}=\mathbf{L}+\mathbf{S}$ also satisfies the relations (1).
From (1) we can prove $[L^2,L_z]=0$. Therefore we know there are simultaneous eigenstates of $L^2$ and $L_z$. We can call these states $|l,m\rangle$ where $l$ labels the $L^2$-eigenvalue and $m$ labels the $L_z$-eigenvalue.
Now we define two operators $L_+$ and $L_-$. $$L_\pm=L_x\pm iL_y \tag{2}$$ These definitions are grabbed from thin air here. So we need to prove that these two operators are actually ladder operators.
It is not hard to calculate the commutators between operators (2) and $L^2$ and $L_z$. Using (1) we find: $$[L^2,L_\pm]=0 \tag{3}$$ $$[L_z,L_\pm]=\pm\hbar L_\pm \tag{4}$$
Now we can look for some properties of the states
$L_+|l,m\rangle$ and $L_-|l,m\rangle$.
From (3) we can prove:
- If $|l,m\rangle$ is an eigenstate of $L^2$, then $L_\pm|l,m\rangle$ is an eigenstate of $L^2$ to the same eigenvalue.
And from (4) we can prove:
- If $|l,m\rangle$ is an eigenstate of $L_z$ to eigenvalue $m\hbar$, then $L_\pm|l,m\rangle$ is an eigenstate of $L_z$ to the eigenvalue $(m\pm 1)\hbar$.
It is these last two properties which show that $L_+$ and $L_-$ are ladder operators. $L_+$ steps up the $L_z$ eigenvalue, and $L_-$ steps down the $L_z$ eigenvalue. $$L_\pm|l,m\rangle \propto |l,m\pm 1\rangle \tag{5}$$