Let $(A^g)_a = (\frac{1}{c} \Phi^g, (A^g)_i)$ has units $[\frac{m}{s}]$.

$(F^g)_{ab} = 2 A^g_{[b,a]}$
$\frac{1}{c} (E^g)_i = (F^g)_{i0} = 2 (A^g)_{[0,i]} = (A^g)_{0,i} - (A^g)_{i,0} = \frac{1}{c} (\Phi^g)_{,i} - \frac{1}{c} (A^g)_{i,t}$ has units $[\frac{1}{s}]$
so $(E^g)_i$ has units $[\frac{m}{s^2}]$
$(B^g)_i = \frac{1}{2} {\epsilon_i}^{jk} (F^g)_{jk} = {\epsilon_i}^{jk} (A^g)_{[k,j]} = {\epsilon_i}^{jk} (A^g)_{k,j}$
so $(B^g)_i$ has units $[\frac{1}{s}]$

$(F^g)_{ab} = \left[\begin{matrix} 0 & -\frac{1}{c} (E^g)_j \\ \frac{1}{c} (E^g)_i & {\epsilon_{ij}}^k (B^g)_k \end{matrix}\right]$ has units $[\frac{1}{s}]$

$(F^g)^{ab} = \left[\begin{matrix} 0 & \frac{1}{c} (E^g)_j \\ -\frac{1}{c} (E^g)_i & {\epsilon_{ij}}^k (B^g)_k \end{matrix}\right]$

${{(F^g)}^{ab}}_{,b} = \mu_g (J^g)^a$ has units $[\frac{1}{m s}]$
${{(F^g)}^{ab}}_{,b} = \frac{4 \pi G}{c^2} (J^g)^a$
so $(J^g)^a$ has units $[\frac{kg}{m^2 \cdot s}]$

${(F^g)^{ab}}_{,b} = \mu_g (J^g)^a$ gives us...
${(F^g)^{0u}}_{,u} = \mu_g (J^g)^0$
${(F^g)^{0u}}_{,u} = \frac{1}{c} (E^g)_{j,j}$
substitute $\frac{1}{c} (E^g)_i = \frac{1}{c} (\Phi^g)_{,i} - \frac{1}{c} (A^g)_{i,t}$
${(F^g)^{0u}}_{,u} = \frac{1}{c} \Phi^g_{,jj} - \frac{1}{c} (A^g)_{j,tj}$
using the Lorentz gauge $(A^g)_{j,j} = 0$ we get
${(F^g)^{0u}}_{,u} = \frac{1}{c} (\Phi^g)_{,jj}$
using $\Delta \Phi^g = -4 \pi \frac{G}{c^2} T_{00}$
we get ${(F^g)^{0u}}_{,u} = -4 \pi \frac{G}{c^3} T_{00}$
and ${(F^g)^{0u}}_{,u} = \mu_g (J^g)^0$
so $\mu_g (J^g)^0 = -4 \pi \frac{G}{c^3} T_{00}$
so $(J^g)_0 = \frac{1}{c} T_{00}$
so $\frac{1}{c} (E^g)_{j,j} = \mu_g (J^g)^0 = -4 \pi \frac{G}{c^3} T_{00}$


${(F^g)^{iu}}_{,u} = \mu_g (J^g)^i$ gives us
${(F^g)^{iu}}_{,u} = -\frac{1}{c^2} (E^g)_{i,t} + {\epsilon_i}^{jk} (B^g)_{k,j} = \mu_g (J^g)^i$
substitute $\frac{1}{c} (E^g)_i = \frac{1}{c} (\Phi^g)_{,i} - \frac{1}{c} (A^g)_{i,t}$ and $(B^g)_i = {\epsilon_i}^{jk} (A^g)_{k,j}$
${(F^g)^{iu}}_{,u} = -(\frac{1}{c^2} \Phi^g_{,it} - \frac{1}{c^2} (A^g)_{i,tt}) + {\epsilon_{ij}}^k {\epsilon_k}^{lm} (A^g)_{m,lj} $
using $(A^g)_{i,tt} = 0$ (or am I thinking of $\bar{h}_{ab,tt} = 0$ ...)
${(F^g)^{iu}}_{,u} = -\frac{1}{c^2} (\Phi^g)_{,it} + (\delta_i^l \delta_j^m - \delta_i^m \delta_j^l) (A^g)_{m,lj} $
${(F^g)^{iu}}_{,u} = -\frac{1}{c^2} (\Phi^g)_{,it} + (A^g)_{j,ij} - (A^g)_{i,jj} $
using the Lorentz gauge $(\Phi^g)_{i,i} = (A^g)_{i,ij} = 0$
${(F^g)^{iu}}_{,u} = -\frac{1}{c^2} (\Phi^g)_{,it} - (A^g)_{i,jj} $
and using $-(\Phi^g)_{,it} = 0$ (why?) and $(A^g)_{i,jj} = -\frac{4 \pi G}{c^3} T_{0i}$
${(F^g)^{iu}}_{,u} = \frac{4 \pi G}{c^3} T_{0i}$
so $\mu_g (J^g)^i = \frac{4 \pi G}{c^3} T_{0i}$
so $(J^g)^i = \frac{1}{c} T_{0i}$


Then the dual:

$(\star F^g)^{ab} = \left[\begin{matrix} 0 & (B^g)_j \\ -(B^g)_i & -\frac{1}{c} {\epsilon_{ij}}^k (E^g)_k \end{matrix}\right]$

Then ${(\star F^g)^{ab}}_{,b} = 0$ gives us...
$(B^g)_{i,i} = 0$
$(B^g)_{i,t} + {\epsilon_i}^{jk} (E^g)_{k,j} = 0$

If we wanted to rewrite the matter variables in terms of stress-energy (so we can consider stress-energy sources other than matter, such as electromagnetism ...):
${(E^g)^i}_{,i} = -4 \pi \frac{G}{c^2} T_{00} = -\frac{1}{c^2 \epsilon_g} T_{00}$
${(B^g)^i}_{,i} = 0$
$(B^g)_{i,t} + {\epsilon_i}^{jk} (E^g)_{k,j} = 0$
$(E^g)_{i,t} - c^2 {\epsilon_i}^{jk} (B^g)_{k,j} = c^2 \mu_g (J^g)^i = \frac{1}{c \epsilon_g} T_{0i} $

Rewrite this in terms of $D^g = \epsilon_g E^g$ which has units $[ \frac{kg}{m^2} ]$
${(D^g)^i}_{,i} = -\frac{1}{c^2} T_{00}$
${(B^g)^i}_{,i} = 0$
$(B^g)_{i,t} + (\epsilon_g)^{-1} {\epsilon_i}^{jk} (D^g)_{k,j} = 0$
$(D^g)_{i,t} - (\mu_g)^{-1} {\epsilon_i}^{jk} (B^g)_{k,j} = \frac{1}{c} T_{0i} $

As a linear system:
$\partial_t \left[ \begin{matrix} (D^g)_i \\ (B^g)_i \end{matrix} \right] + \left[ \begin{matrix} 0 & -\frac{1}{\mu_g} {\epsilon_i}^{jk} \\ \frac{1}{\epsilon_g} {\epsilon_i}^{jk} & 0 \end{matrix} \right] \partial_j \left[ \begin{matrix} (D^g)_k \\ (B^g)_k \end{matrix} \right] = \left[ \begin{matrix} \frac{1}{c} T_{0i} \\ 0 \end{matrix} \right]$

Another thing to keep an eye on is the fact that, in order to solve the Poisson equation, we assume $\partial_t^2 h_{ab} = 0$
I guess we are technically just approximating it to be incredibly low.