To answer your title question, the speed of electricity does not become the speed of light. But many "messages" or, more correctly, transient effects that set up the final steady state flow do indeed travel at the speed of light in the medium your circuit is steeped in
Think of your circuit at the very instant you close the switch. The switch changes a boundary condition of the electromagnetic field, suddenly (let's say for simplcity, instantaneously) imparting a potential difference between the two conductors at the battery end, whereas the potential differnce just before the switch close was nought.
Let $+z$ be the direction along the conductors. For simplicity we think of your conductors as a waveguide whose cross section is translationally invariant in the $z$ direction.
If you solve this boundary value problem given that the initial Potential Difference (PD) was nought but is now suddenly +V at one end of the conductor, the solution is that the potential difference now travels down the conductors as a freespace electromagnetic wave, progressively raising the PD from 0 to +V at points of increasing $z$ as the wave travels.
The speed of this wave is not simple: the waveguide comprising the conductors with freespace between them have a system of eigenmodes with different speeds that depend on the cross sectional geometry. But the lowest order mode, the TEM mode (see footnote) for a two conductor system, indeed runs at speed $c$ if the conductors are steeped in free space. The disturbance comprises eigenmodes that travel at speeds $c$ and less than $c$.
As this disturbance travels, the EM field acts on the conduction electrons in the wires. They begin to move. They can't move at $c$ as you rightly point out because they have nonzero rest mass. As they accelerate, they generate their own radiation: new electromagnetic waves that travel down the conductors as a system of modes with speeds $c$ and less.
Meanwhile, the freespace wave reaches the resistor. If the resistor is not matched to the waveguide characteristic impedance, a reflected wave begins to come back, comprising eigenmodes at speeds travelling $c$ and less.
And all these waves bouncing back and forth move the electrons which radiate their own reaction fields. We get, very fleetingly, a hugely complicated system of waves bouncing back and forth,coming from both the original closing of the switch as well as radiation from the accelerated electrons.
Eventually these transients settle down with the electrons moving uniformly through the conductor at their drift velocity set by the potential difference $V$. This is MUCH slower than $c$. There are no more bouncing waves: the electromagnetic field is now static and the electrons drift in their boring old constant speed pattern around the circuit.
But, because the initial transient includes the TEM eigenmode that travels at $c$, the first motion of electrons at the resistor end of the system indeed begin to move a time $L/c$ after the switch is closed, where $L$ is the $z$-direction distance along the conductors from the battery to the resistor.
Footnote: TEM ("Transverse Electromagnetic") modes arise in two conductor waveguides, i.e. those with a froward and backward "return" path, i.e. a system that can be thought of as a closed circuit. For conductors in freespace, they always have speed $c$ and, curiously, the cross sectional EM field configuration is the same as an electrostatic / magneto static configuration, but this cross sectional static variation is subject to an aplitude change that moves as a wave, as i explain in my answer here and also here.