(1-2) The Sudoku property (a.k.a. Latin square property) of group tables follows from the group axioms, so I don't think there is any intuition to be gleaned. As for the symmetric identity property, this follows from commutativity of inverses. Say $g_r$ and $g_c$ are group elements appearing in row $r$ and column $c$, respectively. To say that the identity $e$ appears in the $r^{th}$ row and $c^{th}$ column means that $g_r g_c = e$. But note that we have $g_c g_r = e$ as well. Thus, $e$ also appears on the $c^{th}$ row and $r^{th}$ column.
(3) As far as I know, associativity cannot easily be seen from a group table in general. The question is only asking for a group table from a set of $3$ elements, which is why it is feasible to construct such a group table being asked for. Did you read Light's associativity test?
(4) The question indeed doesn't mention closure, but I think that is meant to be implied. This is just a simple matter of writing some group element in each row / column.
EDIT: We know that $a \neq b \neq e$, otherwise the group would not have order $3$. Note that the table is symmetric about the diagonal except for $ab \neq ba$. We want to use this to show that something goes wrong (i.e. associativity fails). Since $a = ba$, we have that $\begin{align} a\color{magenta}b & = (ba)\color{magenta}b
\\ b & = \end{align}$.
Also, $b = ab$ implies $\begin{align} \color{darkorange}{b}b & = \color{darkorange}{b}(ab) \\ e & =\end{align}$. But $\color{darkorange}{b}(ab) = e \neq b = (ba)\color{magenta}b$, which shows that associativity fails.
I agree with your answer.
Completing the Cayley table of order $6$ (if $pq=e$, it must be $qp=e$, etc.), you get a non commutative ($pr\ne rp$) magma, with identity element $e$.
Each element of it has unique inverse (each one is inverse of itself, but $p$ and $q$ mutually inverses).
Finally, if you relabel $e,p,q,r,s,t$ as $1,2,3,4,5,6$, you get the table
$$\begin{array}{c|cccccc}
\cdot & 1 & 2 & 3 & 4 & 5 & 6\\
\hline
1 & \color{green}{1} & \color{red}{2} & \color{red}{3} & \color{blue}{4} & \color{blue}{5} & \color{blue}{6}\\
2 & \color{red}{2} & \color{red}{3} & \color{green}{1} & \color{blue}{5} & \color{blue}{6} & \color{blue}{4}\\
3 & \color{red}{3} & \color{green}{1} & \color{red}{2} & \color{blue}{6} & \color{blue}{4} & \color{blue}{5}\\
4 & \color{blue}{4} & \color{blue}{6} & \color{blue}{5} & \color{green}{1} & \color{red}{3} & \color{red}{2}\\
5 & \color{blue}{5} & \color{blue}{4} & \color{blue}{6} & \color{red}{2} & \color{green}{1} & \color{red}{3}\\
6 & \color{blue}{6} & \color{blue}{5} & \color{blue}{4} & \color{red}{3} & \color{red}{2} & \color{green}{1}
\end{array}$$
that you can check to be associative by using this brute-force Matlab script.
On the other hand, the above is a quasigroup since the Cayley table is a latin square and an associative quasigroup has identity element too.
Best Answer
With the translation $e=0$, $a=1$, $b=3$, and $c=2$, we can recognize that our table is the addition table modulo $4$. More formally, the structure $M$ with the given multiplication table is isomorphic to the additive group $\mathbb{Z}_4$, via the mapping $\varphi$ that takes $e$ to $0$, $a$ to $1$, $b$ to $3$, and $c$ to $2$. The fact that the table is a group table then follows from the standard fact that $\mathbb{Z}_4$ is a group.