You can use [itex] W = \int F.ds[/itex], where F is the tension force and ds is the infinitesimal extension. Since the tension force varies from 0 to T during the extension of 2e, the integral works out to (T/2)(2e) = Te. Or if you don't like integrals, the wire is a spring with force constant k = T/2e (per Hookes law F=kx), so you can use energy methods [itex]W = \Delta PE = 1/2kx^2 = (T/(4e))(4e^2) = Te [/itex], same result.
The forces must be equal and opposite for equilibrium. Internally, the applied force T, at one end, must be equal to the internal force, T, in the rope, for equilibrium
extension = FL/AE, where F = the internal tension force and A is the wire cross section and E is the elasticity modulus (note that k=AE/L) , if that is what you mean
its strain energy increases due to the increase of its PE = 1/2kx^2
When the wire obeys hookes law in its elastic range, ignore any cross section change, which only becomes significant in the plastic range beyond yield when the wire will 'neck' down
as long as e is such that the wire still obeys Hooke's law, then the stored energy is valid. Since e is on the order of .001 L, I guess you are right that e^2 must be small