It's important to realize Poynting's theorem, stated in terms of the fields, is valid regardless of what you call $u = \frac{1}{2}\vec E\cdot\vec D + \frac{1}{2}\vec H\cdot\vec B$. In other words, you don't have to assume that this is the electromagnetic energy density (or even use or define the concept of energy density) to derive Poynting's theorem in the form
$$-\frac{\partial}{\partial t}\left({\frac{1}{2}\vec E\cdot\vec D + \frac{1}{2}\vec H\cdot\vec B}\right)=\vec\nabla\cdot(\vec E \times \vec H)+\vec J\cdot\vec E.$$ To put it yet another way, Poynting's theorem doesn't say that energy is distributed according to this density. Recall for instance that in electrostatics we sometimes view energy as being stored in the charges as opposed to the field.
That said, if you do accept $\frac{1}{2}\vec E\cdot\vec D + \frac{1}{2}\vec H\cdot\vec B$ as energy density, then Poynting's theorem becomes a simple and elegant statement of local conservation of electromagnetic energy, with the Poynting vector as the energy flux. If nothing else, this strongly motivates viewing $\frac{1}{2}\vec E\cdot\vec D + \frac{1}{2}\vec H\cdot\vec B$ as the energy density. It should be mentioned however that there are other choices of energy density and Poynting vector that turn Poynting's theorem into a continuity equation for energy density, unless you impose certain constraints on these choices.
To borrow from an earlier answer, Griffiths writes in Introduction to Electrodynamics:
Where is the energy stored? Equations 2.43 and 2.45 offer two different
ways of calculating the same thing. The first is an integral over the charge
distribution; the second is an integral over the field. These can involve completely
different regions. For instance, in the case of the spherical shell (Ex. 2.9) the
charge is confined to the surface, whereas the electric field is everywhere outside
this surface. Where is the energy, then? Is it stored in the field, as Eq. 2.45 seems
to suggest, or is it stored in the charge, as Eq. 2.43 implies? At the present stage
this is simply an unanswerable question: I can tell you what the total energy is,
and I can provide you with several different ways to compute it, but it is
impertinent to worry about where the energy is located. In the context of radiation theory (Chapter 11) it is useful (and in general relativity it is essential) to regard the
energy as stored in the field, with a density
$$\frac{\epsilon_0}{2}E^2 = \text{energy per unit volume.}$$
But in electrostatics one could just as well say it is stored in the charge, with a
density $\frac{1}{2}\rho V$. The difference is purely a matter of bookkeeping.
In short, it's possible that there is no proof of the kind you are looking for, nor am I aware of experimental evidence for $\frac{1}{2}\vec E\cdot\vec D + \frac{1}{2}\vec H\cdot\vec B$ as the energy density other than the observation that general relativity seems to work well.