ADM decomposition:
$g_{ab} = \overset{a(i)\downarrow b(j)\rightarrow}{ \left[ \matrix{ -\alpha^2 + \beta_k \beta^k & \beta_j \\ \beta_i & \gamma_{ij} } \right]}$
$g^{ab} = \overset{a(i)\downarrow b(j)\rightarrow}{ \left[ \matrix{ -1/\alpha^2 & \beta^j /\alpha^2 \\ \beta^i /\alpha^2 & \gamma^{ij} - \beta^i \beta^j /\alpha^2 } \right]}$

adding in the EM potential...
Let $\alpha^2 = \tilde{\alpha}^2 + \gamma^{kl} A_k A_l - 2 A_t$
Let $\beta_i = \tilde{\beta}_i + A_i$
Notice that $\beta^i$ and $\tilde{\beta}^i$ are naturally contravariant while $A_i$ is naturally covariant.
I'm raising and lowering spatial indexes with respect to the spatial metric $\gamma_{ij}$ (just like in all those numerical relativity text books).
This means $A^i = \gamma^{ij} A_j$ versus $A^i$ as a subset of $A^u$, which is $(-A_t + A_j \beta^j) \beta^i / \alpha^2 + A_j \gamma^{ij}$.
Notice the difference between the two, $(-A_t + A_j \beta^j) \beta^i / \alpha^2$, is zero if the shift is zero.

$g_{ab} = a(i)\downarrow \overset{b(j)\rightarrow}{ \left[ \matrix{ -\tilde{\alpha}^2 - \gamma^{kl} A_k A_l + 2 A_t + (\tilde{\beta}_k + A_k) (\tilde{\beta}^k + A^k) & \tilde{\beta}_j + A_j \\ \tilde{\beta}_i + A_i & \gamma_{ij} } \right]}$
$ = a(i)\downarrow \overset{b(j)\rightarrow}{ \left[ \matrix{ -\alpha^2 - \gamma^{kl} A_k A_l + 2 A_t + (\gamma_{kl} \tilde{\beta}^l + A_k) (\tilde{\beta}^k + \gamma^{km} A_m) & \gamma_{jk} \tilde{\beta}^k + A_j \\ \gamma_{ik} \tilde{\beta}^k + A_i & \gamma_{ij} } \right]}$
$ = a(i)\downarrow \overset{b(j)\rightarrow}{ \left[ \matrix{ -\alpha^2 - \gamma^{kl} A_k A_l + 2 A_t + \gamma_{kl} \tilde{\beta}^l \tilde{\beta}^k + \gamma^{kl} A_k A_l + 2 A_k \tilde{\beta}^k & \gamma_{jk} \tilde{\beta}^k + A_j \\ \gamma_{ik} \tilde{\beta}^k + A_i & \gamma_{ij} } \right]}$
$ = \left[\matrix{ -\alpha^2 + \gamma_{kl} \tilde{\beta}^l \tilde{\beta}^k & \gamma_{jk} \tilde{\beta}^k \\ \gamma_{ik} \tilde{\beta}^k & \gamma_{ij} }\right] + \left[\matrix{ 2 A_t & A_j \\ A_i & 0 }\right] + \left[\matrix{ 2 A_k \tilde{\beta}^k & 0 \\ 0 & 0 }\right]$
$= \tilde{g}_{ab} + 2 A_{(a} \delta_{b)t} + (g^*)_{ab}$

So $\Gamma_{abc} = \tilde{\Gamma}_{abc} + (\Gamma^A)_{abc} + (\Gamma^*)_{abc}$.
Now $\tilde{\Gamma}_{abc}$ produces the typical gravitational force.
$(\Gamma^A)_{abc}$ produces forces in the same form as the Lorentz force (see my 'flat space + EM potential' worksheet for how)
and $(\Gamma^*)_{abc}$ is extra, hopefully only applying to the time component, or something else equally insignificant.

How would you incorporate charge?
Aside from the Kaluza-Klein extra-dimension trick...
$\dot{u}^a g_{ai} = \tilde{\Gamma}_{ibc} u^b u^c + \frac{q}{m} (\Gamma^A)_{ibc} u^b u^c + (\Gamma^*)_{ibc} u^b u^c$
$= \tilde{\Gamma}_{ibc} u^b u^c + \frac{q}{m} (E_i (u^t)^2 + \epsilon_{ijk} B_k u^t u^j) + (\Gamma^*)_{ibc} u^b u^c$
...you would need to modify the spacetime geometry itself in order to produce the correction of proportions of charge to mass...
...which I can almost do by absorbing that term into $u^t$...
But I think this is beside the point. Look at the gravitomagnetics papers.
They talk about equivalences between the $g_{tu}$ and $A_u$, and dissect gravity accordingly.
That is essentially what this $A_u$ is -- not an electromagnetic $A$, but instead a gravitomagnetic $A$.

Hmm, another thought on the gravitomagnetic papers.
They assume that -- because electric force comes about from charge, which is the Laplacian of potential ($\phi = A_t$),
that matter force should come about from density, the Laplacian of matter potential (represented along diagonal of $g_{ab}$, namely $\phi_g = g_{tt}$)
However the gravitational force is not proportional to the particles own mass at all (unlike the electric force, which is proportional to the particle's own charge).

Now for inverses:
$g^{ab} = \overset{a(i)\downarrow b(j)\rightarrow}{ \left[ \matrix{ -1/\alpha^2 & (\tilde{\beta}^j + A^j) /\alpha^2 \\ (\tilde{\beta}^i + A^i) /\alpha^2 & \gamma^{ij} - (\tilde{\beta}^i + A^i) (\tilde{\beta}^j + A^j) /\alpha^2 } \right]}$
Looks ugly.
Calculating the geodesic will be much more ugly.
Luckily I've already done this (with my own symbolic tensor package) for the ADM, and just have to replace $\beta^i \rightarrow \beta^i + A^i$

$-{\Gamma^i}_{tt} = \frac{1}{2} \alpha^{-2} \beta^i \beta^j \beta^k {\gamma_{jk}}_{,t} - \frac{1}{2} \alpha^{-2} \beta^i \beta^j \beta^k \beta^l \gamma_{kl,j} - \alpha^{-2} \beta^i \beta^j {\beta^k}_{,j} \beta^l \gamma_{kl} + \alpha^{-1} \alpha_{,j} \beta^i \beta^j + \alpha^{-1} \alpha_{,t} \beta^i - \beta^k \gamma^{ij} \gamma_{jk,t} + \frac{1}{2} \beta^k \beta^l \gamma^{ij} \gamma_{kl,j} + {\beta^k}_{,j} \beta^l \gamma^{ij} \gamma_{kl} - {\beta^i}_{,t} - \alpha \alpha_{,j} \gamma^{ij} $

$-{\Gamma^i}_{mt} = \frac{1}{2} ( \alpha^{-2} \beta^i \beta^j \gamma_{jm,t} - \alpha^{-2} \beta^i \beta^j \beta^k \gamma_{km,j} - \alpha^{-2} \beta^i \beta^j {\beta^k}_{,j} \gamma_{mk} - \alpha^{-2} \beta^i \beta^j {\beta^k}_{,m} \gamma_{jk} + 2 \alpha^{-1} \alpha_{,m} \beta^i - \gamma^{ij} \gamma_{jm,t} - {\beta^i}_{,m} + {\beta^k}_{,j} \gamma^{ij} \gamma_{mk} + \beta^k \gamma^{ij} \gamma_{mk,j} - \beta^k \gamma^{ij} \gamma_{jk,m} )$

$-{\Gamma^i}_{mn} = \frac{1}{2} ( \alpha^{-2} \beta^i \gamma_{mn,t} - \alpha^{-2} \beta^i \beta^j \gamma_{mn,j} - \alpha^{-2} \beta^i {\beta^k}_{,n} \gamma_{mk} - \alpha^{-2} \beta^i {\beta^k}_{,m} \gamma_{nk} - \gamma^{ij} \gamma_{jm,n} - \gamma^{ij} \gamma_{jn,m} + \gamma^{ij} \gamma_{mn,j} )$

Now who knows what I do with this.

Next comes the Maxwell stuff.
Linearized gravitomagnetics does this in terms of the time x spacetime part of - not the $g_{ab}$ (not even the $h_{ab}$ perturbed part) but instead the trace-reversed $\bar{h}_{ab} = h_{ab} - \frac{1}{2} \eta_{ab} {h^c}_c$.
Trace-reversed, meaning $tr(\bar{h}_{ab}) = \eta^{ab} \bar{h}_{ab} = \eta^{ab} h_{ab} - \frac{1}{2} \eta^{ab} \eta_{ab} \eta^{cd} h_{cd} = \eta^{ab} h_{ab} - 2 \eta^{cd} h_{cd} = -tr(h_{ab})$
I wonder what the curved-space metric looks like trace-reversed...
$\bar{g}_{ab} = g_{ab} - \frac{1}{2} g_{ab} g^{cd} g_{cd} = -g_{ab}$. Hmm, so obvious.
Next comes forming Maxwell equations using $\bar{h}_{tu}$ as the equivalent of the electromagnetic 4-potential vector.
I guess that's the same as using $-g_{tu}$ in curved space.
...and the ADM Riemann curvature tensor (will look ugly):
...and the ADM Ricci curvature tensor (will look ugly):
...and the ADM Gaussian curvature (will look ugly):
...and the ADM Gravitational curvature (will look ugly):