It doesn't really mean anything bad at all, in spite of the confused answers one sometimes receives about this matter. The big point is: there is no very direct physical meaning to applying an operator $H$ or $Q$, even if it is an observable, to a wave function $\psi$, even if $\psi$ is in the domain of $H$ or $Q$. (See also Operator versus Linear Transformation, https://physics.stackexchange.com/a/18933/6432 .
In my opinion, the only operator with very direct physical meaning is the time evolution operator $e^{iHt}$ for some $t$, or for all $t$, and as you seem to realise, by the Stone-von Neumann theorem, this exists even for discontinuous $\psi$ even for unbounded $H$.
But, also this is my opinion, the exact definition of Hilbert space is not that important and many mathematical physicists who worry (too much) about whether discontinuous wave functions, which are obviously outside the domain of $H$, are physical or not, manage to formulate Quantum Mechanics just fine on a dense subspace of a Hilbert space. Other, with the opposite worry, again, in my opinion, worrying too much about something which isn't really that bad, talk about rigged Hilbert spaces or nuclear spaces in order to somehow include infinite norm states and exclude non-differentiable wave functions....see Sudbery, Quantum Mechanics and the Particles of Nature. for this.
There is a mathematical reason for thinking that the exact choice of what domain or what space, whether the Hilbert space, or a subspace of smooth wavefunctions inside the Hilbert Space, or the Schwartz space of rapidly decreasing and smooth functions, or an extension of the Hilbert space to include some dual objects such as distributions, to think of as the domain of these operators, is....unimportant because no matter what space you choose, you get the same physics, and that reason is a theorem of Wilfrid Schmid, Henryk Hecht, and Dragan Milicic, or at least somebody or other, which says that if you have a semi-simple Lie group operating on the space, (if the QM is going to made relativistic you eventually have to assume the Lorentz group acts) and if the representation has a finite composition series (this excludes quantum fields), then the algebraic structure of that rep. is independent of which space you consider (within broad limits). Earlier versions of such results were proved by Nelson, Garding, and Harish-Chandra, and gave a very pleasant surprise to Hermann Weyl and everyone else involved at the time...
Now very concretely, even if $\psi$ is not in the domain of $H$, or any other observable $Q$, it is still true that the Hilbert space has a Hilbert basis of eigenstates of $H$ and hence even if $\psi$ is horrible and discontinuous and everything bad, it still holds that
$$\psi = \lim_{n\rightarrow \infty}\sum_{i=0}^{i=n} c_i v_i$$
where $v_i$ is the normalised eigenstate of $H$ (or $Q$) with eigenvalue, say, $\lambda_i$, and $c_i$ are complex numbers, the so-called Fourier--Bessel coefficients of $\psi$, and the convergence is not pointwise but in the L$^2$ norm. Now notice: each finite sum
$$\sum_{i=0}^{i=n} c_i v_i$$ is an analytic function, if $H$ is hypo-elliptic, as is often true, e.g., the harmonic oscillator, and is at any rate smooth and in the domain of $H$.
And it is still true, by the axioms of QM, that the probability that $H$ (or $Q$) will, if measured, take the value of $\lambda_i$, is $\vert c_i \vert ^2$ whether or not $\psi$ is in the domain of $H$ (or $Q$).
Pedagogically, there is this widespread confusion that an observable, since it is an operator, ought to be applied to a function since it is an operator, but this is just a naive confusion. If anything should be applied to the wave function as an operator, it is the exponential of $iH$, which is always bounded. To repeat: just because $H$ is an operator, and $\psi$ is a function, doesn't mean you should apply $H$ to $\psi$. Although when you can, that may be a useful shortcut, it is not necessary to ever do it, and the axioms of QM, when stated carefully, never ask you to apply $H$ to $\psi$. What they ask you to do is, for the unitary time evolution to apply the exponential of $iH$, and for the Born rule probabilities, expand $\psi$ to get its Fourier--Bessel coefficients. The sloppy way of thinking, which one often sees, works fine for many simple QM problems, but leads to people asking precisely this OP, precisely since it is sloppy. The careful axiomatisation states things the way I formulated them.
There's more to it, and the deeper content it encodes is related to symmetries and their associated conserved quantities. Let us start with the classical theory, to see that this is indeed already present there. In Classical Mechanics we may formulate our theory in the Hamiltonian Formalism. In that case we have a phase space $(\Gamma,\Omega)$ where $\Gamma$ is a space, which in basic mechanics courses is usually described as the space of pairs $(q^i,p_i)$ of position and momenta, and where $\Omega$ is an object called sympletic form.
The sympletic form gives rise to an operation among functions on $\Gamma$ called the Poisson bracket $\{,\}$. The Poisson bracket between position and momenta obey $$\{q^i,p_j\}=\delta^i_{\phantom i j}\tag{1}\label{ccr}.$$
Now, you might be aware of a result known as Noether's theorem which puts in correspondence symmetries and conservation laws. In the Hamiltonian Formalism it can be phrased as follows. For a given symmetry we have a function in $\Gamma$, called its Hamiltonian charge $Q$, which has the property that $$\{Q,f\}=-\delta_Q f\tag{2}$$
where $\delta_Q$ is the variation of the observable according to the symmetry corresponding to $Q$.
Now let us consider translations. Consider a translation by $\epsilon^i$ so that the coordinates get transformed as $q^i\to q^i+\epsilon^i$. We will have $\delta q^i = \epsilon^i$. In that regard, observe that if we define $Q = \epsilon^i p_i$ we have $$\{ Q,q^i\}=\epsilon^j\{p_j,q^i\}=-\epsilon^i = -\delta_Q q^i\tag{3}.$$
Observe that (1) has been used in the second equality. What this tells is that (1) is the statement that momentum is the generator of translations, or else that momentum is the Hamiltonian charge associated to translations. In particular, momentum in the $i$-th direction generates translations in the $i$-th direction, that is the content of (1).
This then naturally generalizes to Quantum Mechanics. And it is not so surprising that it happens, since we know that the correspondence principle gives the quantization rule $[] \leftrightarrow i\{\}$. In that setting, the Canonical Commutation Relations are just saying that momentum should be the generator of translations.
Obviously, the whole analysis of symmetries that I have outlined above in Classical Mechanics can be made in a self-contained manner in Quantum Mechanics. I only did it in Classical Mechanics to show you that there is a classical version of the story, which may be easier to understand first.
In summary, commutation relations often encode symmetry statements and their associated charges, and the CCR is just one example of that.
Best Answer
Whenever you have two representations $(V_1,\pi_1)$ and $(V_2,\pi_2)$ of a group or algebra and an isomorphism $\phi : V_1 \to V_2$, then for any $g$ in the group or algebra that isomorphism must send eigenvectors of $\pi_1(g)$ to eigenvectors of $\pi_2(g)$ with the same eigenvalues by definition of an isomorphism of representations (this is the "intertwiner" part).
So when the eigenstates of $\pi_i(g)$ are non-degenerate and a basis of $V_i$, this already uniquely fixes the isomorphism because for each eigenvector there is only a single possible target. There is a slight subtlety here because $x$ and $p$ only have generalized eigenstates, but the same idea still works: Any $v_1\in V_1$ has a representation as $$ v_1 = \int f(x)\lvert x\rangle_1\mathrm{d}x$$ where the $\lvert x\rangle_1$ are eigenstates of $\pi_1(x)$ and this gets sent to $$ \phi(v_1) = \int f(x)\lvert x\rangle_2\mathrm{d}x,$$ i.e. you're just matching wavefunctions. Another way to say this is that you can always construct an isomorphism to the default representation on $L^2(\mathbb{R})$ where $x$ is multiplication by mapping any state $\lvert \psi\rangle$ to its position wavefunction $\psi(x) = \langle x\vert \psi\rangle$, and then concatenating these isomorphisms gives you isomorphisms between any two representations of the CCR.
Note that the above requires that you already know the isomorphism $\phi$ should exist. Your symmetric representation is a representation of the CCR, but in order for us to know that it should be isomorphic to the standard representation on $L^2(\mathbb{R},\mathrm{d}x)$ via Stone-von Neumann you also need to know that it is irreducible. But this is not the case here: You have four operators $x,\partial_x,p,\partial_p$, and so besides your $X = x+\frac{1}{2}\mathrm{i}\partial_p$ and $P = p - \frac{1}{2}\mathrm{i}\partial_x$, there are also the linearly-independent operators $X' = x-\frac{1}{2}\mathrm{i}\partial_p$ and $P' = p + \frac{1}{2}\mathrm{i}\partial_x$ which also fulfill $[X',P'] = [x-\mathrm{i}\partial_p, p + \mathrm{i}\partial_x] = \mathrm{i}$.
So what you have here is a representation of a two-dimensional CCR algebra, and in particular the generalized eigenstates of $X$ or $X'$ alone are not non-degenerate. For $X,X'$ together, you get non-degenerate eigenstates $\lvert x,x'\rangle$, and then the argument from above gives you maps to wavefunctions $\psi(x,x')$ that allow you to map this to the standard position representation $L^2(\mathbb{R}^2,\mathrm{d}x_1\mathrm{d}x_2)$ of the CCR in two spatial dimensions.