Let's "define" (I put quotes since it's not a definition, but just requiring a property) the operator $a$ such that:
$$[a,a^\dagger]=1$$
then
$$n=a^\dagger a$$
No other assumptions are made such that their definition is linked to the position and momentum operators.
From these assumptions only, is it obvious that there is one unique (up to phase and normalization) ket $|n\rangle$ for each given eigenvalue $n$ thus justifying the notation $|n\rangle$ ?
Not at all! In fact, there is nothing stopping us to having two (or more) vacua $|0;1 \rangle$ and $|0;2\rangle$. By applying creation operators on each of these we can obtain precisely what you are asking $|n;1\rangle\propto(a^\dagger)^{n}|0;1 \rangle$ and $|n;2\rangle\propto(a^\dagger)^{n}|0;2 \rangle$.
What is however "clear" (whatever that word means) is that there is a unique irreducible representation on which $a$ can act. In that representation there is a unique vacuum and all eigenspaces of the number operator are non-degenerate. Furthermore, any other representation is obtained by taking direct sums of these, i.e. adding new vacua, as we did above.
In the specific case of the one-dimensional harmonic oscillator for example, one has to use the particular nature of $a$ as an operator on wave-functions to show that the vacuum is unique. That is not given only by the commutation relations.
