Would the floating discharge terminal cause damage to this oscillator?
Pin 7 just goes to a transistor's collector. There's no way it can be damaged by not connecting it to anything, just don't connect it to a voltage source, without a suitable current limiting resistor.
Note the circuit you've posted won't work quite so well in real life, compared to the simulator program. This is because the SPICE model doesn't accurately simulate the 555's output stage. The simulator shows the output swinging between the negative and positive rails, which will not happen if you built this circuit on a breadboard. In reality, the 555's output stage is not a perfect switch, but two bipolar transistors. The upper transistors are a Darlington pair, configured as a follower and will drop at least 1.2V and the lower one is in common emitter configuration and will pull the output close to 0V.
Refer to the following parts of the datasheet.
Figure 2, page 2: the internal schematic, note the transistor on pin 7 (Q14) and the output stage (Q21 & Q22 on the high side and Q24 on the low side).
Figures 6 to 8, page 6: some graphs showing the low output voltage drop vs current. Note it's near zero, at 1mA.
Figure 9, page 8: a graph showing the high voltage drop vs current. Note at 25
oC it's fairly close to 1.4V, irrespective of the current and very temperature dependant.
https://www.st.com/content/ccc/resource/technical/document/datasheet/ba/0a/d7/6e/7c/db/4e/12/CD00000479.pdf/files/CD00000479.pdf/jcr:content/translations/en.CD00000479.pdfIf you built this circuit, the output will swing between about 7.8V and near 0V. The duty cycle will not be 50%, because the capacitor will take a little longer to charge, than discharge. It will also be quite sensitive to changes in temperature, as the voltage drop of the Darlington part has a negative temperature coefficient. Exercise for the reader to figure out whether the duty cycle will be above, or below 50%.