But I can't seem to understand why the total voltage drop in a circuit
is equal to the voltage of the battery?
This is based on Kirchhoff's voltage law that the algebraic sum of the voltages around a closed loop is zero. It is a requirement for conservation of energy. See below.
If the battery gives electrons this energy, where does it go,
otherwise wouldn't electrons continuously gain energy?
The electrical potential energy given the charge by the battery is either dissipated as heat in resistance (due to collisions between the charge and the particles of the resistance), or stored as energy in the electric and magnetic fields of any capacitance or inductance in the circuit.
Further If we had a circuit with just a battery and wires connecting
the negative and positive terminal, would there be a voltage drop?
What would happen and why?
With the exception of superconductors, all conductors (including wires) have some resistance. The voltage drop $V$ across the wires connecting the positive and negative terminals would simply be, from Ohm's law, $V=IR$, where $R$ is the resistance of the wires.
It should be noted that the voltage drop $V$ across the wires is not the voltage between the battery terminals before the wires are connected, but after. That's because all real batteries have internal resistance which effectively limits the maximum amount of current a battery can deliver. So the current delivered by the battery is
Where $r_b$ is the internal battery resistance and $emf$ is the battery voltage with nothing connected to the terminals.
Hope this helps.