Okay, time to gather all this up in an answer.
What does Griffiths show in his book? Boiled down, he shows that there exists an operator $\hat a$ such that:
The combination of these two properties imply that the Hamiltonian has a set of ladder eigenstates. These eigenstates are defined as follows:
We define $|0\rangle$ to be a state satisfying $\hat{a}|0\rangle=0$.
We define $|n\rangle\equiv\frac{1}{\sqrt{n!}}(\hat{a}^\dagger)^n|0\rangle$
Using the two properties of $\hat a$ above, we can prove that each $|n\rangle$ is an eigenstate $\hat H$ with energy $nA+B$.
We know there exists at least ONE such $\hat{a}$, because Griffiths explicitly writes it down in his book and shows it obeys both of the required properties. The question is, can there exist TWO operators with this property?
Let's say $\bar a$ is an operator, and that $\bar{a}$ obeys both properties. I.e., we have:
We follow the identical procedure as above to develop a new set of ladder states $|\bar{n}\rangle$ with energies $\bar{A}\bar{n}+\bar{B}$. There are three possible cases here: either $A\neq\bar A$, or $B\neq \bar B$, or $A=\bar A$ and $B=\bar B$ but somehow still $a\neq \bar a$. We need to show each of these possibilities are impossible.
The proofs will be proofs by contradiction. In each case, we'll select a state, and show that by acting on the state with some operator, we can construct another state with lower energy. We'll show that this process doesn't terminate (doesn't result in the zero vector) no matter how many times we do it. Thus, if we continue long enough, we get negative-energy states. Since we know the simple harmonic oscillator does not have negative energy states, we'll reach a contradiction.
Let's start with $A\neq \bar A$. Without loss of generality, assume $A>\bar{A}$. Then for some $\bar{n}$, we'll have that $|\bar{n}\rangle$ has an energy that cannot be written as $An+B$. Then, by acting with $\hat{a}$, we can generate a whole ladder of states with lower energies. The state $|\phi_m\rangle\equiv(\hat a)^m|\bar n\rangle$ has energy $E_m=\bar A\bar n+\bar B-Am$. We know that $|\phi_m\rangle$ will never equal $|0\rangle$, because $|\phi_m\rangle$ never has the same energy as $|0\rangle$. But since $|0\rangle$ is the unique vector that satisfies $\hat a |0\rangle=0$, that means this process can never terminate; every $m$ gives a nonzero vector in Hilbert space. By making $m$ large, we can make $|\phi_m\rangle$ have negative energy. But the SHO is strictly positive, so this is impossible. We conclude that we cannot have $A\neq \bar A$.
Now, let's assume $A=\bar{A}$, but $B\neq \bar{B}$. WLOG, assume $B<\bar{B}$. Then $|0\rangle$ has energy $B$, while $|\bar{0}\rangle$ has energy $\bar{B}$. In particular, since $|\bar{0}\rangle$ is the unique state which satisfies $\bar{a}|\bar 0\rangle=0$, acting on $|0\rangle$ with $\bar{a}$ produces states of arbitrarily negative energy. The state $|\phi_m\rangle\equiv(\bar a)^m|0\rangle$ has energy $B-Am$, which can be arbitrarily negative if we pick large $m$. We thus conclude that we can't have $B\neq \bar B$.
Finally, say $A=\bar{A}$ and $B=\bar{B}$. We want to show that $a$ and $\bar{a}$ are essentially the same.
We know the matrix elements of $\hat{a}$ are given by
$$
\langle m|\hat{a}|n\rangle = \sqrt{n}\delta_{n-1,m}
$$
Because $\bar{a}$ generates the same ladder of energies, and the spectrum of $H$ is non-degenerate, $\bar{a}$ connects the same states as $a$, up to phases:
$$
\langle m|\bar{a}|n\rangle = e^{i\theta_n}\sqrt{n}\delta_{n-1,m}
$$
where $\theta_n$ possibly depends on the state $n$.
That's as good as you can do: you ARE allowed to pick some random ladder operator that adds phases to your states as you go up and down the ladder. But that's the only freedom you have. Phases aren't really important to the story, so you should consider the ladder operators essentially unique. In particular, you CAN'T have a ladder operator with a different lowest rung (different $B$), and you CAN'T have a ladder operator with a different spacing (different $A$).
Best Answer
Indeed, as suggested by phase-space quantization, most of these equations are reducible to generalized Laguerre's, the cousins of Hermite. As universally customary, I absorb $\hbar$, M and ω into r,E. Note your E is twice the energy.
Since $r\geq 0$ you don't lose negative values, and you may may redefine $r^2\equiv x$, so that $$ r\partial_r = 2x \partial_x \qquad \Longrightarrow r\partial_r (r\partial_r)= r^2\partial_r^2+ r\partial_r=4(x^2\partial_x^2+x\partial_x), $$ hence your radial equation reduces to $$ \left ( \partial_x^2+ \frac{1}{x}\partial_x +\frac{E-x}{4x} -\frac{m^2}{4x^2} \right ) R(m,E)=0 ~. $$
Now, further define $$ R(m,E)\equiv x^{|m|/2} e^{-x/2} ~ \rho(m,E), $$ to get $$ \partial_x R(m,E)= x^{|m|/2} e^{-x/2} \left (-1/2 +\frac{|m|}{2x} + \partial_x \right )~ \rho(m,E) \\ \partial_x^2 R(m,E)= x^{|m|/2} e^{-x/2} \left (-1/2 +\frac{|m|}{2x} + \partial_x \right )^2~ \rho(m,E), $$ whence the generalized Laguerre equation for non-negative m=|m|, $$ x \partial_x^2\rho(m,E) +\left({m+1} -x\right )\partial_x \rho(m,E)+\frac{1}{2}(E/2-m-1) \rho(m,E)=0~. $$ This equation has well-behaved solutions for non-negative integer $$k=(E/2-m-1)/2\geq 0 ~,$$ to wit, generalized Laguerre (Sonine) polynomials $L^{(m)}_k (x)=x^{-m}(\partial_x -1)^k x^{k+m}/k!$.
Plugging into the factorized solution and the above substitutions nets your eigen-wavefunctions. The ground state is $k=0=m$, ($E=2$ in your conventions), so a radially symmetric Gaussian, $e^{-r^2/2}$.
Again, in your idiosyncratic convention, the degeneracy is E/2.
So, degeneracy 2 for $E=4$ : $m=1$, $k=0$; you may check this is just $r e^{-r^2/2 +i\phi} $. You may choose the $\cos \phi$ and $\sin \phi$ solutions, if you wish, constituting a doublet of the underlying degeneracy group SU(2).