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Every metric coefficient is independent of $t$ and $\varphi$. The corresponding coordinate flows therefore preserve the metric, sosatisfy the Killing equation and are Killing vector fields. They generate stationarity and axial symmetry respectively.

Differentiate $E=-m{K}_a{U}^a+q\Phi$ along the worldline. The symmetric contraction $U^a{U}^b{\nabla }_a{K}_b$ vanishes by the Killing equation. Using the Lorentz force and $K^a{F}_{ab}={\nabla }_b\Phi$, the remaining force term cancels $q{U}^a{\nabla }_a\Phi$. Hence

Killing equation is ${\nabla }_{(a}{K}_{b)}=0$. For the stress-energy current from a Killing vector $J^a=T^a{}_b{K}^b$,The first term vanishes on shell, while the second contracts the symmetric tensor $T^{ab}$ with the antisymmetric derivative selected by Killing's equation. Thus $\boxed{{\nabla }_a{J}^a=0}$.