= Solution
Replace $\varphi$ by $\varphi+\epsilon\eta$, with $\eta$ vanishing on the boundary, and differentiate the action at zero. Integrate the derivative of $\eta$ by parts:
$$
\delta S=\int\left[\frac{\partial\mathcal L}{\partial\varphi}\eta+
\frac{\partial\mathcal L}{\partial\varphi_{,a}}\partial_a\eta\right]d^4x
=\int\left[\frac{\partial\mathcal L}{\partial\varphi}
-\partial_a\frac{\partial\mathcal L}{\partial\varphi_{,a}}\right]\eta\,d^4x.
$$
The coefficient of $\eta$ is the <functional derivative> $\delta S/\delta\varphi$.
For the electromagnetic action, $\delta F_{ab}=\partial_a\delta A_b-\partial_b\delta A_a$. Antisymmetry combines these terms and integration by parts gives
$$
\delta S=\int\left[\mu_0^{-1}\partial_aF^{ab}-J^b\right]\delta A_b\,d^4x.
$$
Thus the inhomogeneous <Maxwell equations> are $\partial_aF^{ab}=\mu_0J^b$. The definition of $F$ also gives the homogeneous equations $\partial_{[a}F_{bc]}=0$, equivalently $\partial_a\widetilde F^{ab}=0$. These are the two four-vector Maxwell systems.
Unrestricted variation of the alternative component-gradient action instead gives $\Box A^b=\mu_0J^b$. This is equivalent to the first Maxwell system only when supplemented by the <Lorenz gauge> $\partial_aA^a=0$, because
$$
\partial_aF^{ab}=\Box A^b-\partial^b(\partial_aA^a).
$$
The alternative action is not by itself gauge-invariant; the additional condition must be supplied consistently, not silently omitted.
Expand the tensor square: $F^{ab}F_{ab}=2[\partial^bA^a\partial_bA_a-\partial^bA^a\partial_aA_b]$. Hence
$$
S-\widehat S=\frac1{2\mu_0}\int\partial^bA^a\partial_aA_b\,d^4x.
$$
Integrating this cross term by parts twice expresses its integral as a boundary term plus $\int(\partial_aA^a)^2\,d^4x$. Under Lorenz gauge it is therefore purely a boundary term. With fixed or vanishing boundary contributions, \b[the two actions have the same variational equations once the additional gauge condition is included]. This is the <Lorenz-gauge reduction of the electromagnetic action>.
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