YES. Arc temperatures exceed 5000'K when the switch opens for a few microseconds, and slowly erodes contact silver content and rises in resistance. Better contacts have a palladium-silver alloy which is more robust at higher temps. Cheap contacts have neither silver nor palladium.
This is why DC relay contacts with inductive loads are de-rated to ~25% of rated resistive loads.
Upon reflection, I think a resistor snubber will work best.
The diode method reduces EMI too much and slows down the flipper action to T=L/Rs
So measure solenoid series resistance (DCR) with a DMM and choose a resistor across the inductor about 50x this value. That will only increase current a bit but sped up solenoid response and reduce arc voltage to V=IR for solenoid current and snubber R value.
It depends on your DC solenoid.
If coil draws 0.1A at 5V, it is 5/0.1=50 Ohms ,
then use a 50x 50=2500 Ohms/=50% tolerance 1/2 W
The diode current or power rating must be similar to the coil so it conducts when the switch opens bypassing or shunting the arc voltage to 1 diode drop to the opposite rail. It is normally from the switch in reverse polarity to the opposite supply rail ( + or Return)
Shown below as High and Low side switch. FYI only.
Otherwise, without a diode, high EMI impulse noise is created where the contacts arc. You can hear this between channels on any AM/SW radio. ( as long as the flipper isn't too loud) hah.
Solution 1N400x x= number dont care about reverse voltage rating number) directly across switch in reverse polarity.
simulate this circuit – Schematic created using CircuitLab
The Power supply current does not spike since the diode acts as a bypass switch to the same DC current when the flipper is activated, so the current simply declines smoothly. But without a diode the supply must absorb any transient arc voltage drop ( which it usually does) and this also creates more radiated EMI. So use any 1A diode.
Plan B on left