Charge on capacitors In the given capacitor arrangement while finding the potential and the charge across the capacitors all the solutions say that the charge on both of the capacitors is same. However there is a battery in between the capacitors.
Can we still assume them to be in series? Or is there any other reason that the charge on both the capacitors is same?
The capacitors were initially uncharged. 
 A: 
However there is a battery in between the capacitors. Can we still assume them to be in series?

Yes, they are still in series. The way to tell if they are in series is to see if the current that goes through one is the same as the current that goes through the other. In this case the current is the same so they are in series.
The charge is the integral of the current. Equal current implies equal integral of current. Since they both started uncharged this implies equal charge. Neither the capacitance nor the batteries affect that, and although you didn’t ask even resistors or inductors on the loop would not change the fact that the capacitors are in series.
A: More information is needed to determine the charges on the two capacitors. They are not guaranteed to be equal, unless they are initially uncharged and then a switch in series with all of the component is closed to complete the circuit. In this scenario, charge may be transferred from one capacitor to the other, but since they are initially uncharged and the same current flows through them, they will store the same amount of charge in their final state.
A: The batteries are shown in opposition, so unless $\mathscr E_1$ ≠ $\mathscr E_2$ no current will flow.  Assuming the difference in voltage is $E = \mathscr E_1-\mathscr E_2$, the circuit resistance is R, the circuit capacitance is $C = \frac{C_1C_2}{C_1+C_2}$, and the circuit is closed (e.g. via switch) at time $t = 0$.
The current will be,
$$i(t) =\frac{dq}{dt} =\frac{E}{R}e^\frac{-t}{RC}$$
So, the charge transferred will be the integral of i(t),
$$Q = \int_0^{\infty}\frac{E}{R}e^\frac{-t}{RC} = -CEe^\frac{-t}{RC}\Biggr|_{0}^{\infty}=CE $$
This shows that the charge transfered is not impacted by the circuit R.  R only affects how long it takes (RC time constant) before the transient is essentially over (e.g. 5 time constants).
So, since the fundamental capacitor formula is Q=CV we can then find the steady-state voltage across the individual capacitors as,
$V_{C1}=\frac{Q}{C_1}=\frac{CE}{C_1}=\frac{\frac{C_1C_2}{C_1+C_2}E}{C_1}$
and,
$V_{C2}=\frac{Q}{C_2}=\frac{CE}{C_2}=\frac{\frac{C_1C_2}{C_1+C_2}E}{C_2}$
and, as a check of KVL,
$E=V_{C1}+V_{C2}$
Further, notice that the ratio of the capacitor voltages is then,
$\frac{V_{C1}}{V_{C2}}=\frac{C_2}{C_1}$
Note: Capacitors in series add like resistors in parallel.
UPDATE to accommodate case where one cap has initial charge at t = 0:
Let $C_2$ have an initial charge, $Q_2(0)$, resulting in a voltage of $VC_2(0)$ in opposite polarity to E,
$$i(t) = \frac{E-VC_2(0)}{R}e^{\frac{-t}{RC}}  $$
So, the additional charge transferred to the circuit after switch closed at time t = 0,
$$Q = \int_0^{\infty}\frac{E-VC_2(0)}{R}e^\frac{-t}{RC} = -C[E-VC_2(0)]e^\frac{-t}{RC}\Biggr|_{0}^{\infty}=C[E-VC_2(0)] $$
So, the total charge on $C_1$ will be Q, and the total charge on $C_2$ will now be,
$$Q_2(new) = Q_2(0) + Q$$
