Don't forget the wire is just a low resistance resistor, when current flows through it there is a difference in potential at the two ends of the wire (ie a voltage drop) that is proportional to the resistance times the current (
Ohms law). So yes there is power dissipated in the wire AND there is a real voltage drop across the wire. The fact that this driven by a solar panel doesn't change the basic physics.
Alternatively you can think of this in terms of power - the power dissipated in a resistive element is equal to voltage drop across the element squared divided by the resistance (
Joule Heating P = V^2/R), so by definition, if there is power dissipated in the wire there MUST be a voltage drop across the wire.
PS Sorry to the OP for taking this tangent, but there appear to be some basic misunderstanding about electricity and circuits that are just going to make things more confusing.