My first answer gives a more "holistic" approach, just arguing on the basis of having valid solutions to Maxwell's equations. However, here is another approach that looks in detail at the co-existing fields and current densities, using the microscopic version of Ohm's law. This demonstrates that the reflected wave is indeed produced by currents near the surface of the conductor and that the fields produced by these currents also cancel out the incident electric field further into the conductor.
My initial disquiet about this explanation was based on the fact that for good conductors, the E-field inside the conductor is proportional to $\sigma^{-1/2}$, where $\sigma$ is the conductivity, implying that the current density was proportional to $\sigma^{1/2}$. Thus despite the possibility of $\sigma$ varying by many orders of magnitude in different (good) conductors, the reflected wave amplitude is always very close to the incident wave amplitude.
The key to resolving this is to realise that as the conductivity gets higher, the electric field in the conductor gets lower, but the resultant current density multiplied by the skin depth is almost constant and exactly that required to produced the reflected (and transmitted) waves.
Take as a set-up and incident wave of the form $\vec{E_i} = E_i \cos(\omega t-kz)\hat{i}$, which is incident normally on a conductive plane at $z=0$, with a high conductivity $\sigma$. Let us assume this leads to a transmitted wave in the conductor and a reflected wave. Let us further assume that the wavelength of the incident field is much larger than any skin depth of the conductor, so that in the region where there is a current density the value of $\vec{E_i} \simeq \vec{E_i}(0,t)$.
If the average current density in the medium is $\vec{J}(\omega t)$ and the effective width of the current carrying region is $\delta$, then the electric field generated by the oscillating current sheet will be$\dagger$
$$ \vec{E_s}(z,t) = -\frac{\mu_0 c \delta}{2} \vec{J}(\omega[t- z/c]) \ \ \ \ \ {\rm for\ \ z>0}$$
$$ \vec{E_s}(z,t) = -\frac{\mu_0 c \delta}{2} \vec{J}(\omega[t+ z/c]) \ \ \ \ \ {\rm for\ \ z<0}$$
where $t - z/c$ represents a retarded time.
The total electric fields in the regions $z>0$ and $z<0$ will be $\vec{E_T} = \vec{E_i} + \vec{E_s}$ and from Ohm's law $\vec{J}(t) = \sigma \vec{E_T}(0,t)$.
At $z=0$, we can use either expression for the field generated by the oscillating current sheet to write
$$\vec{J}(t) = \sigma \left( E_i \cos(\omega t)\hat{i} - \frac{\mu_0 c \delta}{2}\vec{J}(t)\right) $$
$$ \vec{J}(t) = \frac{\sigma E_i\cos(\omega t)}{1 + \sigma \mu_0 c \delta/2}\hat{i}\ . $$
This current density can then be used to find $\vec{E_s}$ at $z>0$ and $z<0$.
$$ \vec{E_s} = -\frac{\mu_0 c \delta}{2} \left( \frac{\sigma E_i\cos(\omega t - kz)}{1 + \sigma \mu_0 c \delta/2} \right)\hat{i} \ \ \ \ \ {\rm for\ \ z>0}$$
$$ \vec{E_s} = -\frac{\mu_0 c \delta}{2} \left( \frac{\sigma E_i\cos(\omega t + kz)}{1 + \sigma \mu_0 c \delta/2} \right)\hat{i} \ \ \ \ \ {\rm for\ \ z<0}$$
These fields add to the incident wave to give the total electric field on either side of the current sheet:
$$ \vec{E_T} = E_i \cos(\omega t - kz) \left[ 1 - \frac{\sigma \mu_0 c \delta}{2 + \sigma \mu_0 c \delta} \right] \hat{i} \ \ \ \ \ {\rm for\ \ z>0}$$
$$ \vec{E_T} = E_i \left[ \cos(\omega t - kz) - \frac{\sigma \mu_0 c \delta}{2 + \sigma \mu_0 c \delta}\cos(\omega t + kz) \right] \hat{i} \ \ \ \ \ {\rm for\ \ z<0}$$
For a very good conductor we can require that $\sigma \mu_0 c \delta \gg 2$, these reduce to
$$\vec{E_T} \simeq 0\ \ \ \ \ {\rm for\ \ z>0}$$
$$\vec{E_T} \simeq 2E_i \sin(kz)\sin(\omega t) \hat{i} \ \ \ \ \ {\rm for\ \ z<0}$$
Thus the fields produced by the oscillating current sheet are exactly those required to cancel the incident electric field on the far side of the conductive boundary (and beyond the current-carrying region) and to produce a standing wave on the incident side of the boundary.
I am still thinking about whether I can refine this treatment to deal with the spatial dependence of the electric fields and current density within the current-carrying region rather than assuming it is narrow.
The solution to the paradox I posed in my initial remarks is that although the current density in the conductor is $\propto \sigma^{1/2}$, what matters for the fields that are produced is the product $J \delta$; and since the skin depth is $\propto \sigma^{-1/2}$, this product is almost constant. This is why the reflected wave has almost the same amplitude, independent of the exact value of $\sigma$ (as long as it is large enough to be considered a good conductor).
$\dagger$ Proving the form of the fields radiating from an oscillating current sheet.
Assume a position $\vec{r}=z\hat{k}$ relative to an origin at some point on the plane $z=0$ and that there is a time-dependent current density $J(t) = J\cos(\omega t) \hat{i}$ in a sheet of thickness $\delta$.
The general solution of the inhomogeneous wave equation is
$$\vec{A}(z,t) = \frac{\mu_0}{4\pi} \int \frac{J\cos[\omega(t-|r-r'|/c)]}{|r-r'|}\ d^3r'\ \hat{i},$$
where $\vec{r'}$ is a position on the current sheet, $|r-r'| = \sqrt{z^2 + r'^2}$. If we also let $\vec{J}\ d^3r' = \delta \vec{J} 2\pi r'\ dr'$ and the solution for the vector potential becomes
$$\vec{A}(z,t) = \frac{\mu_0 \delta J}{4\pi} \int^{\infty}_{r'=0} \frac{\cos[\omega(t - k\sqrt{z^2 + r'^2})]}{\sqrt{z^2 + r'^2}}\ 2\pi r'\ dr'\ \hat{i} .$$
If we let $u = k\sqrt{z^2 + r'^2}$, then $du = k r'\ dr'/\sqrt{z^2 + r'^2}$ and
$$\vec{A}(z,t) = \frac{\mu_0 \delta J}{2k} \int^{\infty}_{u=k|z|} \cos(\omega t - u)\ du\ \hat{i} $$
$$\vec{A}(z,t) = -\frac{\mu_0 \delta J}{2 k} \left[\sin(\omega t-k|z|) - \sin (\infty)\right]\ \hat{i}\ . $$
Taking the time derivative
$$ \vec{E} = -\frac{\partial \vec{A}}{\partial t} = -\frac{\mu_0 \delta J \omega}{2 k}\cos(\omega t - kz)\ \hat{i}\ \ \ \ {\rm for}\ z>0 $$
$$\vec{E} = -\frac{\partial \vec{A}}{\partial t} = -\frac{\mu_0 \delta J \omega}{2 k}\cos(\omega t + kz)\ \hat{i}\ \ \ \ {\rm for}\ z<0\, , $$
where $\omega/k = c$ in vacuum.
