Givens:
A:
$a^2 = a \to a = e.$
$ab = a \to b = e.$
$ab = b \to a = e.$
B:
Every row of a group table must contain each element of the group
exactly once.
Problem:
There is exactly one group on any set of three distinct elements, say the set $\{e, a,
b\}$. Indeed, keeping in mind $A$ and $B$ above, there is only one way of completing
the following table. Do so! You need not prove associativity.$$
\begin{array}{c|lcr}
& e & a & b \\
\hline
e & e & a & b \\
a & a & & \\
b & b & &
\end{array}$$
Given:
$ae = ea = a.$
$be = eb = b.$
So, the group has identity $e.$
Suppose $ab = e$. Then $aa = b.$
Since $b = a^{-1}$, then $ba = e.$ So, $bb = a.$
Both $a$ and $b$ have inverses.
Thus:
$$
\begin{array}{c|lcr}
& e & a & b \\
\hline
e & e & a & b \\
a & a & b & e \\
b & b & e & a
\end{array}$$
Does it makes sense? Also, what's the point of $A$ in the givens?
Best Answer
Your final table is correct, but you haven't excluded the possibility of other tables.
The point of A is that it's one particular way of deciding what the entries of the table have to be (instead of guessing at what they could be). For example you know $a \neq e$, so the first part of A tells you that you can't have $ab = b$. Since the second row must contain every element you get that $ab$ is either $b$ or $e$, but it's not $b$, so it must be $e$.