Conservation of energy in an electrical circuit can be expressed by Ampere's law $$\nabla \times \textbf{B} = \mu_o \textbf{J} + \epsilon_o \mu_o \frac {\partial \textbf{E}} {\partial t}$$ when both sides of the equation are dotted with $\textbf{E}$ and the input power is placed on the left side of the equation while the dissipated power (the output power) is placed on the right hand side of the equation
$$\textbf{E} \cdot (\nabla \times \textbf{B}) - \epsilon_o \mu_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} = \textbf{E} \cdot \mu_o \textbf{J} $$
$$ \textbf{E} \cdot \frac {1} {\mu_o} (\nabla \times \textbf{B}) - \epsilon_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} = \textbf{E} \cdot \textbf{J} $$
$$ \textbf{E} \cdot (\nabla \times \textbf{H}) - \epsilon_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} = \textbf{E} \cdot \textbf{J} $$
The units of all terms are $\left[ \frac {W} {m^3} \right]$, which is energy per unit time per cubic meter.
If we now need to get the Poynting theorem from the above equation then we have to add on both sides of the equation the term $+\textbf{B} \cdot \frac {\partial \textbf{B}} {\partial t}$ which from Faraday's law dotted with $\textbf{B}$ is equal to $-\textbf{B} \cdot(\nabla \times \textbf{E})$
$$\textbf{E} \cdot (\nabla \times \textbf{B}) - \epsilon_o \mu_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} +\textbf{B} \cdot \frac {\partial \textbf{B}} {\partial t} = \textbf {E} \cdot \mu_o \textbf{J} +\textbf{B} \cdot \frac {\partial \textbf{B}} {\partial t}$$
$$ \textbf{E} \cdot (\nabla \times \textbf{H}) - \epsilon_o \mu_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} -\textbf{B} \cdot(\nabla \times \textbf{E}) = \textbf{E} \cdot \mu_o \textbf{J} +\textbf{B} \cdot \frac {\partial \textbf{B}} {\partial t}$$
$$ \textbf{E} \cdot (\nabla \times \textbf{H}) -\textbf{B} \cdot(\nabla \times \textbf{E}) = \epsilon_o \mu_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} +\textbf{B} \cdot \frac {\partial \textbf{B}} {\partial t} + \textbf{E} \cdot \mu_o \textbf{J} $$
$$ -\nabla \cdot \textbf{S} = \epsilon_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} + \frac {\textbf{B}} {\mu_o} \cdot \frac {\partial \textbf{B}} {\partial t} + \textbf{E} \cdot \textbf{J} $$
where $\textbf{S} = \frac {1} {\mu_o} \textbf{E} \times \textbf{B} = \textbf{E} \times \textbf{H}$ $\left[ \frac {W} {m^2} \right]$ is the Poynting's vector. Doesn't the above make the Poynting vector redundant in proving CoE, the latter appearing to follow directly from Ampere's law alone?
Notice, it is not true that
$$ \textbf{E} \cdot (\nabla \times \textbf{H}) - \epsilon_o \textbf{E} \cdot \frac {\partial \textbf{E}} {\partial t} = \textbf{E} \cdot \textbf{J} $$
"just states some relation" between vector fields the way it is not true that
$$ \frac{\partial}{\partial t} \biggl[ \frac{1}{2} \biggl( \epsilon_0 \textbf{E}^2 + \frac{\textbf{B}^2}{\mu_0}\biggr) \biggr] $$
"just states some relation" between vector fields. These state $\textit{change in energy density of electromagnetic field}$. As shown above, obtaining of the Poynting vector is only a result of a mathematical manipulation of an equation which already shows balance of these power densities and therefore the Poynting vector and the theorem connected with it are redundant.