A closed system can not speed itself up, that's the momentum conservation law which is also the key to your problem.
As far as I can see you are implicitly supposing the following three equalities to hold
$$E_i+A=E_f\\p_i=p_f\\m_i=m_f$$
where subscripts $i$ and $f$ stays for the initial (before acceleration) for the final (after acceleration) states respectively and $A$ is the energy stored in your battery. However, all three can not exist simultaneously: if $p_i=p_f$ and $m_i=m_f$ then from $E=\frac{p^2}{2m}$ one obtains $E_i=E_f$.
Thus, in order to increase the energy of the system you need to relax one of these conditions. Let us suppose that $m_i=m_f$ but $p_i\neq p_f$, i.e. the momentum is not conserved. For example, our 'car' interacts with the road via friction force $F_{fr}$. Suppose that the speed of the car was increased with the constant acceleration $a$ from the initial value $v$ to the final value $v+\Delta v$. The work done by the friction force on the car is
$$W_{fr}=F_{fr}\int_0^{\Delta v/a}v(t)dt=F_{fr}\int_0^{\Delta v/a}(v+at)dt=F_{fr}\frac{v\Delta v+\Delta v^2}{a}$$
Since the acceleration is caused only by that friction force we also have $ma=F_{fr}$ and therefore
$$W_{fr}=m(\Delta v^2+v\Delta v)$$
Now consider the reference frame moving with a constant speed $v$. In this reference frame the car have traveled less, and less is the work done by the friction force.
$$W'_{fr}=F_{fr}\int_0^{\Delta v/a}v'(t)dt=F_{fr}\int_0^{\Delta v/a}(at)dt=F_{fr}\frac{\Delta v^2}{a}=\frac{m\Delta v^2}{2}$$
You can see that the difference between these two works $W_{fr}-W'_{fr}=mv\Delta v$ is exactly the difference in variations of kinetic energy calculated in these two frames.
Similarly, one can keep condition $p_i=p_f$ but relax condition $m_i=m_f$ thus considering the case of a jet engine. If done accurately, calculations in this case also yield perfect conservation of energy.
