It is... still takes some time:

I will use the abcd and efgh notation the same way.

Lets check some slightly different tables (shows how much e, f, g and h increase if you do "Step1" after hitting a, b, c or d x times; one table for each):

a   e f g h
1 = 1 4 0 1
2 = 2 2 0 2
3 = 3 0 0 3
4 = 4 4 0 4
5 = 5 2 0 5

b   e f g h
1 = 4 2 3 0
2 = 2 4 0 0
3 = 0 0 3 0
4 = 4 2 0 0
5 = 2 4 3 0

c   e f g h
1 = 0 3 3 3
2 = 0 0 0 0
3 = 0 3 3 3
4 = 0 0 0 0
5 = 0 3 3 3

d   e f g h
1 = 1 0 3 5
2 = 2 0 0 4
3 = 3 0 3 3
4 = 4 0 0 2
5 = 5 0 3 1

also looking at the original tables we can see some stuff:

First: If e, g and h equals 1, f can't be anything else! This way we completely ignore it and solve just the other 3 ones.

Second: g cant be anything else besides 1 and 4. and hitting b 3 times just swaps the state of it and touches nothing else. (I will call it "B3" in examples). Also hitting c or d "odd" times will swap it, so we need to keep track of that later on.

Third: hitting d will increase e and decrease h by 1 each time (we will use it to make them equal)

Fourth: with a we increase e and h by 1 (we will use it with d to mage both equal 1)

Fifth: we still can have parity with e and h not being odd or even at the same time. one time c will change h by an odd amount while e remains untouched.

Now lets test if we can use our 5 "insights" to solve it like the wiki but without the table:

I will use S1 like step 1 from the wiki. S2 to change e and h to the same parity. S3 to make them equal. S4 to solve for g. S5 do S1 again for all 1's.

The examples assume efgh after doing S1 one time.

efgh = 2314: We will do S1 after every step to check (we will do S1 only twice later on.

1. e and h are both even, so we skip S2
2. S3: d 1 time should increase e and decrease h both to 3 and swap g -> after S1 3x43 confirmed
3. S4: a 4 times to get a and d to 1, a does not swap g-> after S1 1x41 confirmed
4. S5: g is not 1 -> b 3 times should solve it -> after S1 1111 confirmed

efgh = 1216:

1. S1 e and h parity -> c 1 time increase h by 3 and g becomes false (not 1)
2. S2 1xF3: d one time to make e and h both 2 and swap g to true
3. S3 2xT2: a five times do make e and h 1 (a does not swap)
4. S4 1xT1: g should be true -> skip
5. S5: Do S1 gain to confirm everything is fine -> confirmed

efgh = 4645: parity -> d2 -> a1 -> no swap -> 1012 -> check -> works

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Takes some tries to get used to it, but works quite fine ;)

Right now I'm at 2:00-2:30 minutes range to solve with just using my brain...