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
$$\tag{1}\frac{\hat{H}}{\hbar\omega} ~:=~ \hat{N}+\nu{\bf 1},
\qquad\qquad \nu\in\mathbb{R},$$
$$\tag{2} \hat{N}~:=~\hat{a}^{\dagger}\hat{a}, $$
$$\tag{3} [\hat{a},\hat{a}^{\dagger}]~=~{\bf 1},
\qquad\qquad[{\bf 1}, \cdot]~=~0.$$
(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,
$$\tag{4} {\cal A}~:=~ \text{associative algebra generated by $\hat{a}$, $\hat{a}^{\dagger}$, and ${\bf 1}$}.$$
The spectrum of a semi-positive operator $\hat{N}=\hat{a}^{\dagger}\hat{a}$ is always non-negative,
$$\tag{5} {\rm Spec}(\hat{N})~\subseteq~ [0,\infty[.$$
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
$$\tag{6} |n_0- 1\rangle,\quad |n_0- 2\rangle, \quad\ldots$$
which however could have zero norm. Since the spectrum ${\rm Spec}(\hat{N})$ is bounded from below, this lowering procedure (6) must stop in finite many steps. There must exists an integer $m\in\mathbb{N}_0$ such that zero-norm occurs
$$\tag{7} \hat{a}|n_0 - m\rangle~=~0.$$
Assume that $m$ is the smallest of such integers. The norm is
$$\tag{8} 0 ~=~ || ~\hat{a}|n_0 - m\rangle ~||^2
~=~ \langle n_0 - m|\hat{N}|n_0 - m\rangle
~=~ ( n_0 - m) \underbrace{||~|n_0 - m\rangle~||^2}_{>0},$$
so the original eigenvalue is an integer
$$\tag{9} n_0 ~=~ m\in\mathbb{N}_0,$$
and eq. (7) becomes
$$\tag{10} \hat{a}|0\rangle ~=~0,\qquad\qquad \langle 0 |0\rangle ~\neq~0.$$
We can next use the raising ladder (creation) operator $\hat{a}^{\dagger}$ repeatedly to define new eigenstates
$$\tag{11} |1\rangle,\quad |2\rangle,\quad \ldots.$$
By a similar norm argument, one may see that this raising procedure (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,
$$\tag{12} || ~\hat{a}^{\dagger}|n\rangle ~||^2
~=~ \langle n|\hat{a}\hat{a}^{\dagger}|n\rangle~=~ \langle n|(\hat{N}+1)|n\rangle
~=~ (n+1) ~\langle n|n\rangle~>~0. $$
So $V$ contains at least one full copy of the standard Fock space. 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).
III) Finally, if $V$ is not irreducible, then $V$ could be a direct sum of several Fock spaces. In the latter case, the ground state energy-level is degenerate.
The expression for $O_1$ does not depend on in which single-particle basis $\{|\nu\rangle\}_{\nu}$ you express it.
For example, you can start from your expression of $O_1$ and then use of the relation $$a_{\nu} \equiv \int \mathrm d x \,\varphi^*_{\nu}(x)\, a(x) \quad ,$$ where $\varphi_{\nu}(x)$ is the single-particle wave function corresponding to the single-particle state $|\nu\rangle$. By making use of the orthonormality of these functions and evaluating the matrix element $\langle x^\prime|p^2/ 2m + U(x)|x\rangle$, which should look familiar from non-relativistic (single-particle) quantum mechanics, you will arrive at the desired expression.
Alternatively, and I think this is what you're trying: Try to insert the completeness relation
$$ \mathbb I = \int \mathrm d x\, |x\rangle \langle x|$$ twice (but with different indices) in the matrix element and then make use of
$$a(x) = \sum\limits_{\nu} \varphi{_\nu}(x)\, a_{\nu} \quad .$$
To see why the kinetic energy is diagonal in the momentum representation, just use $o=p^2/2m$ and express $O_1$ in that basis. However, if $o=p^2/2m + U(x)$, then in general $O_1$ will not be diagonal in the momentum basis.
As an example, let us demonstrate how to change the representation of $O_1$ from an orthonormal single-particle basis $\{|\nu\rangle\}_{\nu}$ to $\{|n\rangle\}_{n}$. To start, we see that
$$O_1 = \sum\limits_{\mu\nu} \langle \mu|o|\nu\rangle \, a^\dagger_{\mu}\, a_{\nu} = \sum\limits_{mn} \sum\limits_{\mu \nu} \langle \mu|m\rangle\langle m|o|n\rangle \langle n|\nu\rangle \,a^\dagger_{\mu}\, a_{\nu} \quad , $$
where we have inserted an identity operator (of the single-particle Hilbert space) $ \displaystyle \mathbb I = \sum\limits_n |n\rangle \langle n|$ twice. Now we have to note that
\begin{align}
a^\dagger_m &= \sum\limits_{\mu} \langle \mu |m\rangle \,a^\dagger_{\mu}\\
a_n &= \sum\limits_{\nu} \langle n |\nu\rangle \,a_{\nu} \quad ,
\end{align}
which eventually yields
$$O_1 = \sum\limits_{mn} \langle m|o|n\rangle \, a^\dagger_m a_n \quad .$$
Best Answer
Your observation is actually correct, in the following sense. Consider the matrix elements of the momentum operator:
\begin{align} \langle x \vert \hat{P} \vert y \rangle &= \int dp \langle x \vert \hat{P} \vert p \rangle \langle p \vert y \rangle \\ &= \int dp ~ p~ \langle x \vert p \rangle \langle p \vert y \rangle \\ &= \int \frac{dp}{2\pi} ~ p~ e^{ip(x-y)} \\ &= i \delta'(x-y), \end{align}
where $\delta'(x-y)$ is the generalized derivative of the Dirac delta. You see then that this operator is almost diagonal, but there is this pesky derivative on the Diarc delta. However, when you multiply this by a wavefunction and integrate you'll find you can pass the derivative to the wavefunction and recover a simple Dirac delta.
The Hamiltonian matrix elements then go as
\begin{align} \langle x \vert \hat{H} \vert y \rangle &\propto \langle x \vert \hat{P}^2 \vert y \rangle \\ &\propto \int \frac{dp}{2\pi} ~ p^2~ e^{ip(x-y)} \\ &\propto i^2 \delta''(x-y). \end{align}
Using this into your last expression:
\begin{align} \delta(x-y)E_n\phi_n(y) &= H(x, y)\phi_n(y)\\ &= \left(\frac{-\delta''(x-y)}{2m} + \frac{m\omega^2}{2} y^2 \delta(x-y)\right) \phi_n(y) \end{align}
Integrating on the loose variable ($x$):
\begin{align} E_n\phi_n(y) &= \int dx \delta(x-y)E_n\phi_n(y) \\ &= \int dx\left(\frac{-\delta''(x-y)}{2m} + \frac{m\omega^2}{2} y^2 \delta(x-y)\right) \phi_n(y) \\ &= \left(\int dx \frac{-\delta''(x-y)}{2m} \phi_n(y) \right) + \frac{m\omega^2}{2} y^2 \phi_n(y) \\ &= \left(\int dx \frac{\delta'(x-y)}{2m} \phi'_n(y) \right) + \frac{m\omega^2}{2} y^2 \phi_n(y) \\ &= \left(\int dx \frac{-\delta(x-y)}{2m} \phi''_n(y) \right) + \frac{m\omega^2}{2} y^2 \phi_n(y) \\ &= \left(\frac{-\partial^2_y}{2m} + \frac{m\omega^2 y^2}{2}\right) \phi_n(y) \end{align}