Probs. 12 & 13 , Sec. 2.3, in Herstein's TOPICS IN ALGEBRA, 2nd ed: Existence of only right-sided identity and right-sided inverses suffice

I. Let $G$ be a nonempty set closed under an associative product, which in addition satisfies:

(a) There exists an $e\in G$ such that $a\cdot e=a$ for all $a\in G$.

(b) Give $a\in G$, there exists an element $a^{-1}\in G$ such that $a\cdot a^{-1}=e$.

Prove that $G$ must be a group under this product.

II. Prove, by an example, that right indentity element and left inverse does not imply that $G$ is group.

My solution:

I. Since $G$ is closed set under an associative product, i.e. if $a,b,c\in G$ then $(a\cdot b)\cdot c=a\cdot (b\cdot c)\in G$. Taking $c=e$ we get $(a\cdot b)\cdot e=a\cdot (b\cdot e)=a\cdot b \in G$. We have shown that $\cdot$ is binary operation.

Since $a\in G$ then $a^{-1}\in G$ and we have the following identities $$a^{-1}=a^{-1}\cdot e=a^{-1}\cdot (a\cdot a^{-1})=(a^{-1}\cdot a)\cdot a^{-1} $$ Then $$ \begin{align} e &= a^{-1}\cdot (a^{-1})^{-1} \\ &= \left( \left( a^{-1}\cdot a \right) \cdot a^{-1} \right) \cdot \left( a^{-1}\right)^{-1} \\ &= \left( a^{-1}\cdot a \right) \cdot \left( a^{-1}\cdot \left( a^{-1} \right)^{-1} \right) \\ &= \left( a^{-1}\cdot a \right) \cdot e \\ &=a^{-1}\cdot a. \end{align} $$ Thus we have shown that $$ a\cdot a^{-1}=a^{-1}\cdot a=e. $$ Then we see that $$ e\cdot a= \left( a\cdot a^{-1} \right) \cdot a = a \cdot \left( a^{-1} \cdot a \right) = a \cdot e = a. $$

We have shown that for this set $G$ and the associative binary operation assumed to be defined on $G$, the properties of the existence of a two-sided identity element in $G$ and the existence in $G$ of a two-sided inverse for each element of $G$ are satisfied. Therefore $G$ is indeed a group.

II. But II indeed is true. Lets take the set $G=\{a,b,e\}$ and define the product $\cdot$ by the following identities: $e\cdot e=a\cdot e=b\cdot e=e$ and $a^{-1}=b, \ b^{-1}=a$ and consider the following multiplication table for our set $G$

$\begin{array}{c | c c c c c} \hline\hline & e & a & b \\ \hline e & e & b & b & \\ a & a & a & e & \\ b & b & e & a & \\ \hline \end{array} $

It's easy to verify that conditions of second problem hold for our $G$, however, $G$ is not group since we can show that $b=a$.

Is my reasoning above correct?

EDIT: Maybe this is a duplicate but I would like to know if my solution is true since I have solved it by myself. Especiaaly I am interested in the solution of the second problem.

You don't need to verify that the operation is defined and associative: that's already given.

What you need to show is that

1. $e$ is a left identity as well as a right identity (the latter condition is given)
2. $a^{-1}a=e$, for every $a\in G$

On the other hand, using $a^{-1}$ may be misleading, but your argument seems good. For the sake of clarity, I'll denote by $b$ and $c$ elements such that $ab=e$ and $bc=e$. Your argument becomes $$ b=be=b(ab)=(ba)b $$ then $$ e=bc=((ba)b)c=(ba)(bc)=(ba)e=ba $$ Therefore $$ ea=(ab)a=a(ba)=ae=a $$ Good work!

The operation you give the Cayley table of does not define a group structure on $\{e,a,b\}$, because $ea=eb$, but $a\ne b$ (this is better than saying that “we can show that $a=b$, which is false at the outset). So long as you verify it is associative, you have your counterexample.