Usually there's two diodes in the circuit going from the battery to the ram. If one is bad it can cause wired issues like this

Test the diodes in circuit but without power. If D1 is a Schottky (barrier) diode then the drop should be around 0.2V. D2 should be a small signal diode and the drop should be 0.4V to 0.6V.

D1 blocks the batteries from powering the rest of the board. D2 blocks the batteries from being charged while there is main power.

You shouldn't get a voltage drop in both directions. Use a DMM with a diode test. Red lead on anode and black lead on cathode.

Check your methodology against a known good diode elsewhere in the system. If you are reading a drop in both directions for any tested diode then there's something wrong with your methodology. If the methodology is sound and both D1 and D2 read as you described, I would replace both of them. Of course, the parts are so cheap you could shotgun replace them. I, generally, don't replace parts unless they test bad or I suspect they are bad.

If D1 is shorted the batteries will try to power the entire board which will drain them out quite fast. Not very typical fail mode, but it happens.

https://www.fluke.com/en-us/learn/blog/digital-multimeters/how-to-test-diodes

"Diode test analysis
A shorted diode has the same voltage drop reading in both directions"

You should get a voltage drop in one direction and then it should be open in the other. A diode only lets current pass in one direction and blocks it in the other; it's like a check valve, but for electricity. I have seen crazy results when measuring in circuit, but then pulling it out of circuit it then tests normally. I know people here say not to shotgun replace stuff, but it's so cheap and easy, I would just replace both and re-test and hopefully be done with it.

You can test this diode in circuit. Battery out. Power off.

When testing diodes, remember black on band is when you should see the voltage drop, 0.2 for this special type of diode. Opposite way should be OL / no reading.

If you get voltage drop with the leads across d1 with red on band, the diode is bad. It's leaking reverse. Silicon diodes usually fail dead short, this is a special type of diode that has low voltage drop may have different fail modes like being reverse leakage.

You can probably get by with a 1n4148, but the ram chip will have roughly 0.2vdv less from the extra voltage drop. Williams used a 1n4148 in the earlier boardset and it worked fine. I'd probably get the right part, but if you want the game running for Xmas, go for it.

For sanity I checked a real sys 11 MPU. On diode test i don't get any reading reverse on D1 as expected. If you are getting a reading there is a leak from somewhere. If it goes away when U25 is removed I guess that would be a clue.

I guess if Q40 shorted it could drain out the batteries through the base and possible still work OK since it just passes the RAM CE to the chip when the CPU is out of reset.

6116 should be a low power version intended for battery backup. This chip is typically hard soldered to the board sine you mention a socket its possible someone stuffed the board with a chip that is not have a lower power shutdown mode. What is printed on your 6116 chip?

Just get with the program and install NVRAM. It never forgets, even if you do.

That chip is the issue as it has 5mA standby current. If the batteries have 3000mAh that is only 600hours of life.
http://www.datasheet-pdf.com/PDF/TMM2016BP-10-Datasheet-Toshiba-784932

Since the RAM is hard soldered in, didnt really expect the wrong part to go in.

The math is about right for two days. CR2032 ~ 250 mAh / 5mA = 50hours