When we define electric potential of a charge, we need to take a
charged body as a reference and then we bring the test charge to
calculate the electric potential.
The reference point (point assigned a potential of zero) need not be a charged body. It can be any point arbitrarily chosen (though generally logically chosen based on the analysis involved). The important thing is there needs to be an electric field between the two points in order for there to be a potential and potential difference. This follows from the definition of electric potential:
The electric potential is the amount of work required to move a unit charge from a reference point (point assigned an electric potential of zero) to a specific point against the direction of an electric field between the two points.
That the potential is based on the work done against the electric field is important because, by convention, the direction of the electric field is taken as the direction of the force that a positive charge would experience if placed in the field. That, in turn, is based on the convention in electrical engineering that current is the flow of positive charge (so called conventional current). This convention was established well before electrical current was understood to primarily consist of the flow of electrons.
Now, the voltage of battery being $V$ would mean that the positive
charges have $V$ energy per coulomb more than the negative charges.
It means that it takes $V$ Joules of positive work to move one coulomb of positive charge from the negative battery terminal to the positive battery terminal. It is positive work because the positive charge has to be moved against the (repulsive) direction of the electric field. Positive work transfers energy to the charge increasing its potential and potential energy, relative to the negative charge.
But how are we being able to define energy and potential without a
reference point in this case?
The reference point is the negative terminal of the battery. Then the potential is zero at the negative terminal and $V$ at the positive terminal making the potential difference $V$. Note that if we assigned the negative terminal a potential of 1 volt, the potential at the positive terminal would be $V+1$, and the potential difference would still be $V$.
The difference between potentials would exist if we calculated
everything with respect to a certain electrical charge.
If by "certain electrical charge" you mean some other charge than at the battery terminals, then yes. The potentials at the battery terminals would be different but the potential difference would be the same.
So, with respect to which charge are we calculating the difference in
electrical energy and potential of the positive and negative charges
in this case?
The negative terminal of the battery, as described above.
With regard to your following comment as to why the negative terminal is considered the reference point:
could you please explain why that's the case? Why is the positive
charge considered to be at a higher potential?Why can't we assign the
higher one to negative charges since the scenario is quite symmetrical
where the words positive and negative don't seem to matter
You're correct about the symmetry. The negative terminal would be at higher potential than the positive terminal if, by convention, the direction of the electric field was chosen as the direction of the force that a negative charge would experience if placed in the field, and current was chosen to be the flow of negative (electron) charge. The convention for current is unfortunate, but we've learned to live with it.
Hope this helps.
