What would a Cayley table of inverse semigroup look like?

As you mentioned yourself, the set of all partial bijections on a set $X$ is an inverse monoid. If you take the set of all partial bijections that are not bijective, you get an inverse semigroup that is not a monoid. For instance, if $X$ is a two-element set, you get the five element Brandt semigroup. It can also be described as the semigroup $$ S = {a, b, ab, ba, 0} $$ generated by $a$ and $b$ under the relations $aba = a$, $bab = b$ and $aa = bb = 0$. Up to the identity, this example was given to you in a comment that you ignored and was deleted by his author. Your example is the same, in a more complicated presentation, since you use a set with $5$ elements instead of just $2$.

Another description of the semigroup is $$ S = left{pmatrix{1&0\0&0}, pmatrix{0&1\0&0}, pmatrix{0&0\1&0}, pmatrix{0&0\0&1}, pmatrix{0&0\0&0} right} $$ a description (again up to the identity) that was also given in a comment that you ignore.

I finally solved my problem! :)

I found a structure and table which is pure inverse semigroup and it table. So solution all other similar questions is this. Behold:

'...1 2 3 4 5

1...1 1 1 1 1

2...1 2 3 1 1

3...1 1 1 2 3

4...1 4 5 1 1

5...1 1 1 4 5

I dont know how to type table :) But: no identity!, Associativity proved by Light's test :), even not commutative, and iverse by definitions aba=a, bab=b. Inverses are: a-a, b-b, c-d, d-c, e-e!!! I am so happy.

This is pure inverse semigroup!