...inside of the housing was some sort of mica wafer assembly wired in series with the +binding post. I would assume that this was some sort of resistor, but it is no longer conducting current.
I've never heard of these but a series resistor would be required for the high voltage.
Applying just one volt runs the meter to 125V, 3 volts and the needle pegs.
So full scale should be at \$ \frac {300V}{125V}1V = 2.4~V \$.
Is there an inline resistor I can choose which can be used to make this meter function at line voltage?
Assuming there is a scale on the other side of the meter and that you want to keep it as a 300 V DC meter we need to calculate the series resistor required.
- Measure the meter coil resistance using a multimeter.
- The full scale current can be calculated using this value and the 2.4 V full-scale (FS) that we calculated above.
$$ I_{FS} = \frac {V_{FS}}{R_{COIL}} = \frac {2.4}{R_{COIL}} $$

simulate this circuit – Schematic created using CircuitLab
Figure 1. Voltage range conversion is as simple as adding a series resistor.
- The series resistor needs to drop the remainder of the voltage at \$I_{FS}\$.
$$ R1 = \frac {300-2.4}{I_{FS}} $$
I'd be interested to know how to wire it more directly w/o a wall wart so that it could read higher voltages.

simulate this circuit
Figure 2. Using a wall-wart to convert the meter to measure AC mains voltage safely.
You just need a transformer with a voltage rating greater than or equal to the highest AC voltage you will be measuring.