I can give this a shot. Anyone smarter than me is welcome to offer corrections.
The first issue, based on your question is how much closer you move the Moon. (I'm assuming that you're not playing with other variations like the Earth's rotation speed or the Moon's mass).
If you want Tides to be 100 times greater, you have to move the Moon 10 times closer. The tidal force is equal to the square of the distance (Radius to Radius, not surface to surface). Putting the Moon 10 times closer would make it enormous in the sky and full moons would be bright enough to read by and the Tidal force would be 100 times greater.
The Tidal Force is basically a bulge in the Earth that stays pointed at the Moon. The land bulges too, not just the oceans.
Optimally, tides require about 1/4 of the circumference of the Earth to have full effect. That's why lakes basically don't have tides. Anything smaller than an ocean doesn't have tides to speak of. Now if you make the tieds 10 times greater, the ratio ocean to large lake would still be the same, but large lake and sea tides (like the Mediterranean) would become much more apparent.
Think of tides as raising the sea level of the entire ocean. Of-course, it's not really like that, but locally, that's essentially what happens.
If the current tide is 50 CM (above average sea level), then imagine, in that area, what it would be like if sea level rose 50 meters. In-land it wouldn't be an issue, but most coastal cities couldn't survive that. 100 times the tides would be the end of virtually all coastal living.
See chart.
Source.
As far as nearly geostationary, that would have 2 effects. The slower movement would create more leveling off, less variation that you see in the chart above, where one part of the Earth can get tides that are amplified can get 2 or 3 times the tides than average, and the Tides would happen more slowly, as you get two high tides every time the Moon is overhead. If the Moon took 6 months between rises, you'd get a high tide every 3 months, but you'd still have the 50 foot tide problem. The geostationary orbit distance of the Moon is pretty close to the first number given, about 36,000 km.
Currently Tides vary quite a bit, but that's because the Sun and Moon both interact with tides, so you get pretty significant tidal variation. You also get variation as the Moon's distance to the Earth changes. If the Moon was 10 times closer, the Sun-tides wouldn't change and would be arguably ignorable (less than 1/2 of 1% variation), where, currently the position of the sun to the Moon is important in tidal calculation.
You'd need to take into account the Moon's orbital period and the earth's rotation to estimate how quickly the tides move around the Earth. Currently the Moon takes 29 days to orbit the Earth, so it's primarily the Earth's rotation (once every 24 hours) that governs the tides. If you bring the Moon into 36,000 or 37,000 km, then the orbital speed is important because every few thousand KM would make a measurable difference.
My answer is pretty general. I can try to touch on some specifics if I missed anything. You could also ask in Astronomy stack exchange. Tides are perhaps closer to Astronomy than Earth Science.
Shouldn't the tidal forces go as the inverse cube of distance, not the
inverse square? Given that and that orbit radius is the only change
It's definitely the inverse square law. I've done enough reading up on tidal forces to be sure of that. The formula also clearly shows it as a product of inverse square too. See here or just google "Tidal force formula". - Google's a handy shortcut for looking up formulas.
But I get what you're saying. Thinking about it gets a little complicated. This website gives a better explanation than I'm likely to. As a rule of thumb, the tidal force is equal to the size of the object in the sky (distance from the center of the Earth, so near by objects don't count), times the density. The Sun and Moon are about the same size in the sky, the Moon is more dense, so the Moon causes greater tides.
Shouldn't such a large change be partially damped by the resulting
increase in basin volume as low-lying areas flood?
I hadn't thought of that, and probably, but I don't think that would affect it a great deal. When a tidal bulge happens, the entire ocean (from the point of view of land) lifts up. Oceans are very large, so unless the land is very flat for a very long distance, I don't think there's going to be a huge difference. Lets say the tidal bulge peaks at 50 meters. The tidal "wavelength" if you will is 10,000 km from peak to troth. 5,000 km from 50 meters to zero. While it's a sign wave, not a linear drop, if we measure it linearly, which would be an exageration, the tidal bulge drops 1 meter every 100 km of distance. Land usually rises quite a bit faster than that along the coastline almost everywhere. There are some low-lands that would permanently become lakes, but mostly, I'd say at least 90%-95%, you can calculate how far the tide would reach based on meters above sea level. If the land stays flat for a long time, you could adjust downwards. That's my best guess anyway. I'm not really an expert.
The bigger issue is the shape of the coastline. Some coastlines focus the tide and effectively increase it. The map above gives a pretty good estimate of that. Basically, when the coastline is convex, the tide is focused and increases and when the coastline is concave, it decreases. The exception being the gulf of Mexico, which has very low tides, despite being a very convex body of water, but the tides are blocked by Florida and many islands. If you slow the orbit of the Moon down significantly, like you said, a moonrise every 6 months, then I think that effect would be largely reduced due to the much slower moving tide. It would have a lot more time to even out.
Could oceanic boundary conditions limit tides non-linearly? Might the
increased tidal mass increase response time and flatten the bulge?
Not crystal clear what you're asking here.
