= Solution
The <Maxwell equations> with vacuum permittivity and permeability give $\rho=\epsilon_0\nabla\cdot E$ and $J=\mu_0^{-1}\nabla\times B-\epsilon_0\partial_tE$. The current identity is the <Ampère-Maxwell equation>, and <Faraday's law> gives $\partial_tB=-\nabla\times E$. With $g=\epsilon_0E\times B$,
$$
f+\partial_tg=\epsilon_0[(\nabla\cdot E)E-E\times(\nabla\times E)]+
\mu_0^{-1}(\nabla\times B)\times B.
$$
The identity $(\nabla\times B)\times B=(B\cdot\nabla)B-\tfrac12\nabla B^2$, and its electric analogue, turn this into the divergence of the <Maxwell stress tensor> $T$. Thus the tensor in the question is its negative:
$$
\boxed{\sigma_{ij}=\epsilon_0\left(\frac12E^2\delta_{ij}-E_iE_j\right)
+\frac1{\mu_0}\left(\frac12B^2\delta_{ij}-B_iB_j\right),\qquad
\partial_tg_i+\partial_j\sigma_{ij}=-f_i.}
$$
The <vector> $g$ is <electromagnetic momentum density>, and $\sigma_{ij}$ is the outward flux of its $i$th component across a surface normal in the $j$ direction. In static conditions the total <force> on matter is $F_i=-\oint\sigma_{ij}n_j\,dS$; generally subtract the time <derivative> of the enclosed <electromagnetic field> <momentum> as well.
For the two infinite sheets, superposition and the <electromagnetic field> jumps give $E=(\sigma/\epsilon_0)e_x$, $B=\mu_0K e_y$ between them and zero outside. Substitution yields exactly
$$
\sigma_{ij}=\frac{\sigma^2}{2\epsilon_0}\operatorname{diag}(-1,1,1)
+\frac{\mu_0K^2}{2}\operatorname{diag}(1,-1,1).
$$
A thin pillbox enclosing the $x=0$ sheet has only its $+x$ face in the interior <electromagnetic field>. Therefore
$$
\boxed{\frac FA=\left(\frac{\sigma^2}{2\epsilon_0}-\frac{\mu_0K^2}{2}\right)e_x.}
$$
Directly, omit the sheet's self-<electromagnetic field>: the other sheet supplies $E_{\rm other}=\sigma e_x/(2\epsilon_0)$ and $B_{\rm other}=\mu_0K e_y/2$. The <Lorentz force> per area is $\sigma E_{\rm other}+K e_z\times B_{\rm other}$, giving the same result. The electric contribution is attractive and the magnetic contribution repulsive.
If $K=\sigma v$ for a single moving charge population, use $\mu_0\epsilon_0=c^{-2}$ to obtain $F/A=\sigma^2(1-v^2/c^2)e_x/(2\epsilon_0)$. For physical $|v|<c$ this is attractive, so \b[motion of these charges cannot make the <force> repulsive]. Independently imposed currents need not satisfy that restriction on $K/\sigma$.
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