= Solution
Set $H(a,b)=|a-b|$ and $Q(a,b)=\operatorname{sgn}(a-b)(f(a)-f(b))$. Both are symmetric in $a,b$. For fixed $(s,y)$ apply the entropy inequality for $u$ at the constant level $k=v(s,y)$ with test $\Phi(t,x,s,y)$, then integrate over $(s,y)$. This produces
$$
\int H(u,v)\Phi_t+Q(u,v)\Phi_x+\int H(u_0(x),v(s,y))\Phi(0,x,s,y)\ge0.
$$
Here the first integral is over both time-space pairs and the second over the three remaining variables. Repeat with $v$, level $u(t,x)$, integrating over $(t,x)$, to obtain the analogous terms with $\Phi_s,\Phi_y$ and initial value $v_0$.
Adding and using symmetry gives
$$
\boxed{\int\bigl[H(u,v)(\Phi_t+\Phi_s)+Q(u,v)(\Phi_x+\Phi_y)\bigr]+B_u+B_v\ge0,}
$$
where $B_u=\int H(u_0(x),v(s,y))\Phi(0,x,s,y)\,ds\,dx\,dy$ and $B_v=\int H(u(t,x),v_0(y))\Phi(t,x,0,y)\,dt\,dx\,dy$. All integrals are finite on the test's compact support because the states are bounded. The levels being inserted are constants relative to the variables in each entropy inequality; there is no differentiation of the merely measurable other solution. <Fubini's theorem> and continuity of $H,Q$ justify the integration. This is <doubling of variables for scalar conservation laws>.
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