The term Cayley table is generally restricted to finite groups. However, it’s certainly possible to generalize the idea. For a group $G$ and an element $a\in G$, the $a$ ‘row’ of the table is essentially just the function $$f_a:G\to G:b\mapsto ab\;,$$ and the $a$ ‘column’ is essentially just the function $$f^a:G\to G:b\mapsto ba\;.$$ If $G$ is countably infinite, you can visualize the Cayley table as an infinite matrix.
Let $G$ be any group, and fix $a\in G$. For each $b\in G$ you have $b=a(a^{-1}b)$, so $b$ appears in row $a$ in column $a^{-1}b$. Similarly, $b=(ba^{-1})a$, so $b$ appears in column $a$ in row $ba^{-1}$. It follows that $b$ appears in every row and column. The cardinality of the group doesn’t matter.
Added: You didn’t ask, but it’s also clear that each element of $G$ appears only once in each row and column: if $ax=ay$ or $xa=ya$, then $x=y$. Thus, each of the maps $f_a$ and $f^a$ for $a\in G$ is a bijection from $G$ onto itself, i.e., a permutation of $G$. The set of all permutations of $G$ is denoted by $\operatorname{Sym}(G)$ and is a group under composition of functions; the maps
$$G\to\operatorname{Sym}(G):a\mapsto f_a$$
and
$$G\to\operatorname{Sym}(G):a\mapsto f^a$$
are isomorphisms of $G$ to subgroups of $\operatorname{Sym}(G)$. This is Cayley’s theorem.
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
None of the properties fail automatically
For any group property, you can always find a Cayley table where there's a duplicate entry and yet that property still holds. Here are examples for each:
Associativity can still hold. $$\begin{array}{c|cc} \times& 0 & 1 \\\hline 0 & 0 & 0\\1 & 0 & 1\end{array}$$
Identity can still hold. (Same example.) $$\begin{array}{c|cc} \times& 0 & 1 \\\hline 0 & 0 & 0\\1 & 0 & 1\end{array}$$
Inverses can still hold. (Here, $a$ and $b$ are inverses of each other.)$$\begin{array}{c|ccc} & e & a & b \\\hline e & e & a & b\\a & a & a & e \\b & b & e & a\end{array}$$
However, either associativity fails or inverses fail.
If there's a duplicate row, then $ab=ac$ for some $b\neq c$. Suppose the operator has inverses and associativity. Then we get $a^{-1}ab = a^{-1}ac$ so that $b=c$— contradicting our assumption that $b\neq c$.
So if there's a duplicate row, the operator can either be associative (as shown above), or have inverses (as shown above), but never both.
For confirmation, note that in the example tables above, #1 is associative but not invertible because of 0, and #3 is invertible but not associative because $(bb)a = aa = a$ but $b(ba) = be = b$.)
Diagram
Groups can't have repeat entries. Therefore, if a table has repeat entries, it's not a group. If it's not a group, then it's not in the green region of this diagram. Visually, you can see that such a table can't be both associative and have inverses at the same time. And you can show that there exist tables with duplicate rows that belong to any other non-green region of this diagram.