Additional condition needed for a binary operation to have inverse, apart from the necessary condition of having a two-sided identity.

Is it needed that if two sided identity exists for a binary operation (on a set), still need to check if inverse exists or not.

I request a reason for the same.

I came across this issue in exercise in Chap. $2$ in ‘Problem C — Operations on a Two-Element Set‘; in the book on Abstract Algebra by Charles Pinter.
The solutions there state that out of four operations (out of $16$), i.e. $O_1, O_6, O_7, O_9$, that have two-sided identity; only two operations , i.e. $O_6, O_9$, have inverse.

I am confused over the discrepancy of having a two-sided identity; yet no inverse. I mean why two-sided identity is not a sufficient condition for inverse to exist wrt a binary operation on a set.
All I know that a two sided identity is needed for a unique inverse for any element in a set wrt a binary operation.

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Edit:
Seek formal reasoning; as say exists for why two sided identity is a must for inverse to exist. To, show it is not sufficient, but only a necessary condition for inverse to exist.
It means that need show that identity finding lacks certain conditions that are needed to find inverse. Or, in other words; identity finding operation is a subset of the Inverse finding operation.

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Edit 2:
In my comment below to @JaapScherphuis, have concluded that the condition (additional) to have inverse is to satisfy $a^2=e$ for each non-identity element in the set. Please vindicate or contradict.

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Edit 3:
My doubt is how is it possible for a binary operation on a set (with order $2$) to have two-sided identity, yet not have $x^2 =e$ for each non-identity element $x$ in the set.
The answer for me is in the tables constructed, as it shows the possibility.
But, would be more satisfied if got answer for such binops on sets with order $ge 3$.