Energy balance in battery thermodynamics
The answer to this problem explains how to arrive at 

$$ dQ_p = - dw_{ele} +  dG + TdS $$

This expression is generally true for electrical work at constant T and p. From it the differential equation in the OP can be derived by taking the time derivative on both sides and making appropriate substitutions. 


  $\frac{\mathrm dG}{\mathrm dt}$ can be expanded with dT:  $\frac{\mathrm dT}{\mathrm dt}S = -\frac{\mathrm dT}{\mathrm dt}\frac{\mathrm dG}{\mathrm dT}$


But remember that you defined 

$G=H-TS$

and you should use that when computing $\frac{\mathrm dG}{\mathrm dt}$. But introducing that just results in a circular argument (a lot of terms cancelling to no use). 


  to get to the equation at the very beginning of this post, it is necessary to set $E_0$ and $dE_0/dT$ constant


That's ok. $E_0$ is the voltage when no current is running (that is, battery is not being depleted, as there is no reaction, and therefore composition is constant over time), and is thus constant with respect to time.


  But this makes no sense to me, since the whole purpose of this calculation is the temperature increase with time during cycling a cell. 


That's tricky and I do not have a complete answer. The equation derived is for a specific T. To solve for the time-dependence of the temperature evidently requires additional steps, presumably incorporating expressions for the heat capacity.