Well, I'm not sure I understood your question so I'm going to write what I think and let's see if it's useful :-)
The algebra $[a,a^\dagger]=1$ is all you need to diagonalise $H$, but this is because what $H$ looks like:
$$
H=\omega a^\dagger a
$$
The important observables, namely $H,P,X$, can be written as polynomials in $a,a^\dagger$:
\begin{aligned}
X&=a+a^\dagger\\
P&=i(a-a^\dagger)\\
H&=\omega a^\dagger a
\end{aligned}
and, of course, we can show that any observable $\mathcal O(P,X)$ can be written as a linear combination of monomials in $a,a^\dagger$.
Now, diagonalising $H$ is the same as solving for the time evolution of operators, because in the basis where $H$ is diagonal time evolution is trivial. But, time evolution is given by the commutator $[\mathcal O,H]$, and using the product rule and the linearity of $[\cdot,\cdot]$, it's easy to see that if we know $[a,a]$, $[a^\dagger,a^\dagger]$ and $[a,a^\dagger]$, we know the commutator of any observable $\mathcal O$ with $H$, that is, we know the time evolution of any observable.
For example,
\begin{aligned}
i\dot X&=[X,H]=\omega[a+a^\dagger,a^\dagger a]=[a,a^\dagger a]+[a^\dagger,a^\dagger a]=\\
&=\omega\left(a^\dagger[a, a]+[a,a^\dagger ]a+a^\dagger[a^\dagger,a]+[a^\dagger,a^\dagger ]a\right)
\end{aligned}
where I only used the algebraic properties of $[\cdot,\cdot]$.
With this, if we know the individual commutators $[a,a]=[a^\dagger,a^\dagger]=0$ and $[a,a^\dagger]=1$ we can write
\begin{aligned}
i\dot X&=\omega(a-a^\dagger)
\end{aligned}
and, by taking a second derivative we get $\ddot X+\omega^2X=0$, that is, we find the explicit form (ODE) of the time evolution of $X(t)$.
Conclusion: the algebra of $a,a^\dagger$ is enough to completely specify the commutator of $H=\omega a^\dagger a$ with any operator $\mathcal O(X,P)$, and therefore it's enough to determine the time evolution of any observable. This in turns means that, once we know, $[a,a]$, $[a^\dagger,a^\dagger]$ and $[a^\dagger,a]$, we know the eigenvalues of $H$.
EDIT
There are two ways to introduce the $a,a^\dagger$ operators.
1) Dirac's way (as can be found on most books on QM): We assume that there exist two operators $X,P$ that we take as fundamental, and define
$$
H=\frac12P^2+\frac12X^2
$$
together with $[X,P]=i$. From this, the usual analysis follows (see e.g., here where they motivate the definition of $a$ and diagonalise $H$).
In this method, all the observables can be written as polynomials in $X$ and $P$, that is, as polynomials in $a,a^\dagger$.
2) Weinberg's method (see Weinberg I. for more details): We assume that there exists a discrete basis $|n\rangle$ $n=0,1,2,\dots$ such that any $\psi$ can be written as $|\psi\rangle= c_n|n\rangle$ (implicit sum). Then we can write
$$
a|n\rangle=|n-1\rangle\quad a^\dagger|n\rangle=|n+1\rangle
$$
up to a normalisation, and this defines the operator $a$ and its commutation relations. With this, we can prove that any operator $\mathcal O$ can be written as
$$
\mathcal O=o_0 \mathbb 1+o_i a^i+o_{ij}a^ia^j+o_{ijk}a^ia^ja^k
$$
where $\{a^i\}=\{a,a^\dagger\}$ and there are implicit sums over repeated indices. The proof of this theorem can be found in W. I, but the meaning is very simple: any operator can be written as a linear combination of $a,a^\dagger$.
In this picture, the operators $a,a^\dagger$ are "fundamental", and we can define, for example, $X=a+a^\dagger$. Now, how do we know that $H\propto a^\dagger a$? well, we don't. But WLOG we can write
$$
H=h_1a+h_1^*a^\dagger+h_2 a^\dagger a+\text{cubic terms}+\dots
$$
but the terms with $h_1$ would make $H$ unbounded (as can be seen by evaluating $\langle 1|H|0\rangle$), so we must take $h_1=0$. This means that $H=\omega a^\dagger a$ plus higher order terms. These higher order terms would make the EoM for $a$ non-linear, which means that we must neglect them if we want a harmonic oscillator (which is linear, by definition).
This analysis shows how we can derive the usual harmonic oscillator if we assume that $a,a^\dagger$ are the fundamental operators. In any case, it should be clear that, whether we treat $a$ as fundamental or derived, the commutator $[a,a^\dagger]$ is all we need to find the eigenvalues of $H$, because diagonalising $H$ is the same as solving the time-evolution, which in turns is given by $[\mathcal O,H]$. As in both 1) and 2) we can write any $\mathcal O$ as a polynomial in $a,a^\dagger$, once we know $[a,a^\dagger]$ we know $[\mathcal O,H]$ for any $\mathcal O$.
