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Use real physical fields in the quadratic energy expressions. The vector identity $\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot (\nabla \times \mathbf{E})-\mathbf{E}\cdot (\nabla \times \mathbf{B})$ and vacuum Maxwell's equations giveIntegrating over a fixed volume and applying the divergence theorem proves Poynting's theorem:The left side is outward energy flux through the closed surface, not merely local intensity.

For a linearly polarized wave with phase $\chi =\mathbf{k}\cdot \mathbf{r}-\omega t$, the real fields give Poynting vector $\mathbf{S}={\epsilon }_{0}c{E}_0^2{\cos }^2\chi \hat{\mathbf{k}}$. Local intensity oscillates at $2\omega$. In the closed-surface integral, the time-independent part cancels because ${\oint }_A\hat{\mathbf{k}}\cdot d\mathbf{A}=0$. The remaining flux generally oscillates at $2\omega$, has zero time average, and may vanish identically for special volume geometries.

For circular polarization, take $\mathbf{E}=E_0({\mathbf{e}}_1\cos \chi +{\mathbf{e}}_2\sin \chi )$ with orthonormal transverse vectors. Then $E^2=E_0^2$ and $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$, so $\mathbf{S}={\epsilon }_{0}c{E}_0^2\hat{\mathbf{k}}$ is spatially and temporally constant. Therefore the closed-surface Poynting flux of a plane wave isThere is still a nonzero steady local energy flow: equal energy enters and leaves every fixed closed volume. Calling the net closed-surface flux a nonzero constant would confuse these two quantities.