Are you sure that jumper exists (my board does not have it) ?
"Active" state is logic state, inputs can have "high" and "low" state, you can set logic state in Mach and tell him how to interpret input state into active state. If you actually do have jumper that can make inputs "high" when left in the air, then you do not need resistor and you do not have to connect +5V to the input since it will be "high" allreaday. If not, try 2.2K resistor betwen input and +5V instead of 10K - I think you'll have to have one resistor for each input you want to use that way (thats how I have it wired, all my inputs are pulled high, someone correct me if I'm wrong).
Then wire your switches and/or probe so they are "N.O." - normaly open. When contact is closed it should bring 0V line to input pin (bringing it "down" into "low" state). Mach should be configured for that input(s) so that active state is low ("Active low").
Now, I have 0VDC and AC ground on the same potential - my PSU has it internaly shorted, but your PSU most probably may not have it, so you'll have to short it yourself so you bring 0VDC to tool. Since your tool is now grounded (probably via router frame) and ground is shorted to 0VDC, it may introduce noise into system, so you may need to use capacitor (0.1uF) between input line and ground to filter it out (capacitor lets AC pass and blocks DC) - i do not need/use it, and I have only read about it (no expirience there).
BTW. re AC ground and 0VDC, 0VDC is reference potential for signal lines from which DC voltages are measured, and AC ground is used for safety, so that if any current appears on metal parts exposed to touch (where it would be hazardous) it would go to ground through that wire instead through you.
For the conversation sake, what would happen if 24VAC present on router case/tool was sent to ground via capacitor ? Does it depend on the current source/magnitude available on that 24VAC, and if so how can that current magnitude be measured ?