I) It depends on how abstract OP wants it to be. Say that we discard any reference to 1D geometry and position and momentum operators $\hat{q}$ and $\hat{p}$. Say that we only know that
$$\frac{\hat{H}}{\hbar\omega} ~:=~ \hat{N}+\nu{\bf 1},
\qquad\qquad \nu\in\mathbb{R},\tag{1}$$
$$ \hat{N}~:=~\hat{a}^{\dagger}\hat{a}, \tag{2}$$
$$ [\hat{a},\hat{a}^{\dagger}]~=~{\bf 1},
\qquad\qquad[{\bf 1}, \cdot]~=~0.\tag{3}$$
(Since we have cut any reference to geometry, there is no longer any reason why $\nu$ should be a half, so we have generalized it to an arbitrary real number $\nu\in\mathbb{R}$.)
II) Next, assume that the physical states live in an inner product space $(V,\langle \cdot,\cdot \rangle )$, and that $V$ form a non-trivial irreducible unitary representation of the Heisenberg algebra,
$$ {\cal A}~:=~ \text{associative algebra generated by $\hat{a}$, $\hat{a}^{\dagger}$, and ${\bf 1}$}.\tag{4}\label{eq:4}$$
The spectrum of a semi-positive operator $\hat{N}=\hat{a}^{\dagger}\hat{a}$ is always non-negative,
$$ {\rm Spec}(\hat{N})~\subseteq~ [0,\infty[.\tag{5}$$
In particular, the spectrum ${\rm Spec}(\hat{N})$ is bounded from below. Since the operator $\hat{N}$ commutes with the Hamiltonian $\hat{H}$, we can use $\hat{N}$ to classify the physical states. Let us sketch how the standard argument goes. Say that $|n_0\rangle\neq 0$ is a normalized eigenstate for $\hat{N}$ with eigenvalue $n_0\in[0,\infty[$. We can use the lowering ladder (annihilation) operator $\hat{a}$ repeatedly to define new eigenstates
$$ |n_0- 1\rangle,\quad |n_0- 2\rangle, \quad\ldots \tag{6}\label{eq:6}$$
which, however, could have zero norm. Since the spectrum ${\rm Spec}(\hat{N})$ is bounded from below, this lowering procedure $\eqref{eq:6}$ must stop in finite many steps. There must exists an integer $m\in\mathbb{N}_0$ such that zero-norm occurs
$$ \hat{a}|n_0 - m\rangle~=~0.\tag{7}\label{eq:7}$$
Assume that $m$ is the smallest of such integers. The norm is
$$\begin{align} 0 ~=~& || ~\hat{a}|n_0 - m\rangle ~||^2 \cr
~=~& \langle n_0 - m|\hat{N}|n_0 - m\rangle \cr
~=~& ( n_0 - m) \underbrace{||~|n_0 - m\rangle~||^2}_{>0},
\end{align}\tag{8}$$
so the original eigenvalue is an integer
$$ n_0 ~=~ m\in\mathbb{N}_0,\tag{9}$$
and eq. $\eqref{eq:7}$ becomes
$$ \hat{a}|0\rangle ~=~0,\qquad\qquad \langle 0 |0\rangle ~\neq~0.\tag{10}$$
We can next use the raising ladder (creation) operator $\hat{a}^{\dagger}$ repeatedly to define new eigenstates
$$ |1\rangle,\quad |2\rangle,\quad \ldots.\tag{11}\label{eq:11}$$
By a similar norm argument, one may see that this raising procedure $\eqref{eq:11}$ cannot eventually create a zero-norm state, and hence it goes on forever/doesn't stop. Inductively, at stage $n\in\mathbb{N}_0$, the norm remains non-zero,
$$ \begin{align} || ~\hat{a}^{\dagger}|n\rangle ~||^2
~=~& \langle n|\hat{a}\hat{a}^{\dagger}|n\rangle\cr
~=~& \langle n|(\hat{N}+1)|n\rangle\cr
~=~& (n+1) ~\langle n|n\rangle~>~0. \end{align}\tag{12}$$
So $V$ contains at least one full copy of a standard Fock space
$${\rm span}_{\mathbb{C}}\left\{|n\rangle \mid n\in\mathbb{N}_0\right\}. \tag{13}\label{eq:13}$$
Notice that the Fock space $\eqref{eq:13}$ is invariant under the action of the Heisenberg algebra $\eqref{eq:4}$, i.e. it is a representation thereof.
On the other hand, by the irreducibility assumption, the vector space $V$ cannot be bigger, and $V$ is, hence, just a standard Fock space (up to isomorphism). In this case, the normalized eigenstates $|n\rangle$ are unique up to phase factors.
III) Finally, if $V$ is not irreducible, then the operator $\hat{N}$ has an eigenstate outside the Fock space $\eqref{eq:13}$. We can then repeat the argument in section II to find a linearly independent copy of a standard Fock space inside $V$, i.e. $V$ becomes a direct sum of several Fock spaces. In this latter case, the ground state energy-level is degenerate.
