For the raising operator case: We know that $\hat{a}_+^\dagger = \hat{a}_-$
(Don't forget that you operator acting on your conjugate is daggered)
Therefore $\langle\psi_n$|$\hat{a}_- \hat{a}_+$|$\psi_n\rangle$ = $|c_n|^2$ $\langle\psi_{n+1}|\psi_{n+1}\rangle$ = $|c_n|^2$
But $[\hat{a}_-, \hat{a}_+]=1$ and when you expand the commutator out you get $\hat{a}_- \hat{a}_+ = \hat{a}_+\hat{a}_- + 1$
So substituting in we get:
$\langle\psi_n$|$\hat{a}_+ \hat{a}_-$ + 1|$\psi_n\rangle$ = $|c_n|^2$
We also know that $\hat{a}_+ \hat{a}_-=\hat{N}$ which is the number operator with eigenvalues equal to $n$.
Therefore $\langle\psi_n$|$\hat{N}+1$|$\psi_n\rangle$ = $|c_n|^2$ which leads to:
$n + 1$ = $|c_n|^2$ and finally we get: $\sqrt{n+1} = c_n$
With a very similar manner you derive the lowering operator eigenvalue, and also the Hamiltonian! Just remember that $\hat{H} = \hbar\omega(\hat{a}_+ \hat{a}_- + \frac{1}{2})$
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
This is a fantastic question! Let's get started.
I will assume that we already have defined the ladder operators $a$ and $a^{\dagger}$ and have defined a "ground state" $|0\rangle$ (we still have not proved it is the ground state) such that $a|0\rangle=0$. We will also assume that we already know that the Hamiltonian of the Harmonic oscillator can be written in the form
$$H=\hbar\omega\left(a^{\dagger}a+\frac{1}{2}\right).$$
(Note that the ground state $|0\rangle$ is trivially an eigenstate with $E_0=\hbar\omega/2$.) Finally, I will assume that we have already shown the commutation relations of the ladder operators. Namely,
$$[a,a^{\dagger}]=1.$$
With this, we have enough for a proof.
We can define a state $|n\rangle$ (let's forget about normalization for now) as
$$|n\rangle=(a^{\dagger})^n|0\rangle,$$
where $n$ is a nonnegative integer. The state $|n\rangle$ is an eigenstate of the Hamiltonian with energy $E_n=\hbar\omega(n+1/2)$. We wish to show that the set $\{|n\rangle\}_{n\in\mathbb{Z}^+}$ are all of the possible normalizable eigenstates of the Hamiltonian.
Recall that in the position representation, if we have a potential $V(x)$, then we cannot have a normalizable eigenstate $|\psi\rangle$ whose energy satisfies $E_{\psi}\leq\min V(x)$. That is, we can not have an energy less than the minimum potential energy of the system (ie the kinetic energy must be positive).
Now, we finish off with a proof by contradiction. Consider an eigenstate $|\psi\rangle$ whose energy is given by $E_{\psi}=\hbar\omega(n+1/2+\epsilon)$, with $\epsilon\in(0,1)$. Such a state would essentially describe any of the "other" states that $H$ could permit. Now, consider the state $|\psi^{(1)}\rangle=a|\psi\rangle$. By the commutator algebra, it is not hard to show that $|\psi^{(1)}\rangle$ has energy
$$E_{\psi^{(1)}}=\hbar\omega\left((n-1)+\frac{1}{2}+\epsilon\right).$$
Now, we can induct and define a state $|\psi^{(m)}\rangle\equiv(a^m)|\psi\rangle$. Clearly, its energy is given by
$$E_{\psi^{(m)}}=\hbar\omega\left((n-m)+\frac{1}{2}+\epsilon\right).$$
Thus, unless this process terminates at some point (that is, $a|\psi^{(m)}\rangle=0$ for some $m$), we can achieve an arbitrarily low energy. However, this process could never terminate, since the ground state $|0\rangle$ is unique (it's defined in terms of a position operator and a single derivative operator, so $a|0\rangle=0$ simply defines a first order differential equation in position space) and has energy $\hbar\omega/2$, this cannot be achieved for any $\epsilon$ in the given range. Thus, no such state $|\psi\rangle$ can occur. Similarly, we cannot have a state with energy $E_{\psi}\in(0,\hbar\omega/2)$ by the same logic.
Thus, we have (very rigorously) shown that the only normalizable of $H$ are those with energy $\hbar\omega(n+1/2)$, which are uniquely made from the action of ladder operators on the ground state.
I hope this helped!
(TL;DR -- If another state did exist, it would have an energy not of the form of those given by ladder operators. However, acting on this state many times with $a$ would produce an arbitrarily low energy, and is thus such a state could not exist.)