TLDR: the brief answer to your title question is: No, the Poynting vector represents the flow of energy in all cases including a DC circuit. There is nothing magical about DC that changes the meaning of the Poynting vector.
Neither field is delivering energy outside the wire. The Poynting vector represents potential energy outside the circuit
Unfortunately, this is incorrect, and seems to be a result of a misunderstanding of Poynting’s theorem. Poynting’s theorem can be written: $$\frac{\partial}{\partial t} u + \nabla \cdot \vec S + \vec E \cdot \vec J=0$$ where the Poynting vector is $\vec S = \vec E \times \vec H$. In this equation $u$ is energy density with SI units of $\text{J/m^3}$ and $\vec S$ is an energy flux in SI units of $\text{W/m^2}$
In Poynting’s theorem the amount of energy in the fields at any point is given by $u$ and the work done on matter at any point is $\vec E \cdot \vec J$. Both of those quantities deal with energy at a given point, either energy in the field or energy leaving the field as work on matter. The Poynting vector is the only term that represents the movement of energy through the fields from one point to another. It does not even have the right units to represent a storage of potential energy. It is a flow of energy.
The current inside the wire is delivering the energy to the resistor. Interaction between the current electrons and the resistor results in the transfer of energy.
This is correct but incomplete. The $\vec E \cdot \vec J$ term does result in the transfer of energy from the EM fields to matter, but it does not provide a movement of energy from one location to another. In particular, like $u$, it is a scalar not a vector, so it cannot define a direction for energy to flow. So the question arises, where does that energy come from. That is the role of the Poynting vector. In Poynting’s theorem it is the only term that is both a vector and has the right units for energy flow.
So, the correct understanding of Poynting’s theorem is that the potential energy density in the field at each point is given by $u$ and the interaction with matter is given by $\vec E \cdot \vec J$ at each point while the flow of energy from one place to another is given by $\vec S$
Note that Poynting’s theorem is derived from Maxwell’s equations. Any system which follows Maxwell’s equations follows Poynting’s theorem. The frequency does not matter. Specifically for this question, it applies for static fields and DC circuits which do obey Maxwell's equations.
Edit responding to your newly posted references and comments:
Note that your first reference is from a known predatory publisher AASCIT. You should be highly skeptical of anything found in such journals. They do not actually do peer-review. They simply publish anything submitted, provided the author pays their publication fee. Their business model is based on deception and preys on unsuspecting young researchers and unsuspecting readers, both of whom are tricked into believing that their peer review process is real. In any case, Maxwell's equations trump this article, and Poynting's theorem follows directly from Maxwell's equations.
The reason that this paper was published in a predatory journal instead of a real peer-reviewed journal is that the analysis of their experimental setup is vastly insufficient for supporting their claim. Extraordinary claims require extraordinary evidence. When you seek to experimentally disprove a theory you MUST calculate the exact prediction of the theory for your experimental setup according to the rules of that theory. Such a calculation using the Poynting vector for his "shielded" setup was not even attempted. Without that calculation it is impossible to compare the experimental results to the Poynting vector's theoretical prediction in order to assert a failure of the theory. There is simply no way to make their conclusion based on their analysis.
Your second reference agrees with what I have been describing. Additionally, here is another reference that I like, which quantitatively shows the flow of energy in a coaxial cable: Manoj K. Harbola, "Energy flow from a battery to other circuit elements: Role of surface charges" American Journal of Physics 78, 1203 (2010). Note in particular figures 4, 5, and 6 showing respectively the E-field, the direction of the Poynting vector, and the magnitude of the longitudinal component of the Poynting vector. Figure 6 directly supports your second reference which qualitatively found "that electromagnetic energy flows ... mainly in the vicinity of the wires (and not inside them)".
Regarding the electric field in the wire. In an ideal conductor it is identically 0, and it increases as the resistance of the conductor increases. An increased conductor resistance is associated with increased energy dissipation and reduced energy transmission. So it is immediately clear without even looking at Poynting's theorem that the E-field inside the wire is not associated with energy transfer through the wire but rather with energy dissipation in the wire.
If we use Poynting's theorem inside the wire we gain further insight: the direction of this E-field is longitudinally along the wire, and the magnetic field is circumferential even inside the wire. This results, per Poynting's theorem, in an energy flux into the wire, not along the wire. This energy flux is exactly equal to the $\vec E \cdot \vec J$ energy dissipation in the wire.
Best Answer
If the increase of current is fast enough, the induced electric field will be substantial and then it will contribute to net field energy. However, these experiments (increasing electric current through a coil) are usually meant to be done in a quasistatic way, in which case the induced electric field is very small, so its field energy is negligible. If we changed the current very fast (by increasing voltage on the coil very fast), there would be jump in induced electric field, a radiation-like disturbance and then we are in the realm of EM radiation theory, which is complicated.
The induced electric field is proportional to $dI/dt$, so it is also proportional to acceleration of the mobile charges in the wires. If the field is due to single accelerated charge, it is also called "acceleration field", so we can think of induced field as a sum of many acceleration fields.
This induced field is very different from the Coulomb field - it is solenoidal (non-conservative, curly). Its changes (due to change in $dI/dt$) do propagate out of the system to infinity with speed of light and decay as $1/r$, so in that sense it is always a radiation field. But in usual (low frequency) AC circuits, $dI/dt$ is low enough or geometry is radiation-inhibiting enough that this radiation field carries negligible energy away from the system and we can pretend there is no radiation, only exchange of energy between system components. But if the geometry is radiative or frequency of the current oscillations is high enough, the energy loss may be substantial and then the radiation character must be taken into account, then it is necessary to take the more correct view and talk about the acceleration field being radiative.