The small current observed at the lamp before the electron wave had the time to travel through the wire was energy emitted by the lamp. That energy had to come from the battery (there is no other power source), and it didn't travel through the wire (because there wasn't time), so it traveled through the air.
You are ignoring the capacitors formed by the transmission line wires one on each side of the lamp.
When you are charging two capacitors in series from a source you will not say that energy passed through capacitors but energy went in the capacitor and was stored. While energy was pushed into capacitors it traveled through the wires. You can can not have electric current through a dielectric in this case 1m of air and with no current you have no energy.
This graph I made to show power at the supply and power at the lamp if understood will explain everything.
The switch is only closed for 30ns then open and stays open
With green is power in mW delivered by the supply and with magenta is power used by the lamp.
The area under the graphs represents the energy.
Notice that energy from supply was delivered only during those 30ns that switch was closed circuit and as soon as circuit was broken no more energy is supplied by the source. The total energy delivered by the source in those initial 30ns is 9.16nJ as you can see in the grey window on the left last value.
The lamp was still getting energy well after the switch was closed. Most of the energy arrives at the lamp after the switch was closed and first large chunk exactly after the energy had the time to travel the entire distance through wire.
In the end 7.68nJ of energy where delivered to the lamp and the difference is all accounted by energy lost as heat in the wire's
There is no energy radiated away all energy is accounted for and all will eventually be radiated as photons form the lamp and infrared photons from the wire as heat is lost to environment.