A, B and C only
When the plates of a parallel plate capacitor connected to a battery are moved closer:
$\bullet$ The capacitance $C$ of the capacitor increases because $C = \frac{\epsilon_0 A}{d}$, where $d$ is the separation between the plates, and $d$ decreases.
$\bullet$ Since the capacitor is connected to a battery, the voltage $V$ remains constant. The increased capacitance leads to an increase in the charge $Q = CV$.
$\bullet$ The product of charge and voltage, $Q \cdot V$, increases as the charge $Q$ increases.
Thus, statements A, C, and E are correct.