The "ring" question asks "how many ways are there to put 6 people in a ring", where two "ways" are identical if each individual has the same left and right neighbours.
The "table" question asks "how many ways are there to put 6 people at a table", where two "ways" are identical if each individual has the same left and right neighbours and the same place at the table.
It is easy to see that since the possible rotations are 6, the answer to the first question will be 6 times the answer to the second question, and indeed 6! = 6 x 5!.
Concerning the "necklace" problem, the difference from the "ring" problem is that you cannot put people upside down without noticing - while you can to so with a necklace, obtaining a pattern which is symmetric. So the number of combination is halved, because two symmetric combinations are considered as one.
Given $n$ people, where $n$ is even, you can choose the first pair of people ${n \choose 2}$ ways, where ${n \choose 2}=\frac{n!}{2!(n-2)!}$. The next pair can be chosen in ${n-2\choose 2}$ ways, etc... The end result will be $\frac{n}{2}$ pairs which can be arranged in $\left(\frac{n}{2}\right)!$ ways. So there are $$\frac{{n \choose 2}{n-2 \choose 2}\dots{2 \choose 2}}{\left(\frac{n}{2}\right)!}=\frac{n!}{\left(\frac{n}{2}\right)!\,2^{\left(\frac{n}{2}\right)}}$$ ways to arrange all $n$ people into sets of pairs.
So for 8 people there are $\frac{8!}{4!\,2^{4}}=105$ possible sets of pairs.
Now the question remains, how many sets are there that do not contain any pairs from the original set?
Let $$f(x)=\frac{x!}{\left(\frac{x}{2}\right)!\,2^{\frac{x}{2}}}$$
If $S=\left\{p_{1}, p_{2}, ..., p_{\frac{n}{2}}\right\}$ is our original set of pairs and $P_{k}$ is the set of all sets containing $p_{k}$, where $1\le k\le\frac{n}{2}$ then by the inclusion-exclusion principal the number of sets not containing any of the original pair is:
$$f(n)-\left(|P_{1}|+|P_{2}|+\dots+|P_{\frac{n}{2}}|\right)+\left(|P_{1}\cap P_{2}|+(|P_{1}\cap P_{3}|+\dots\right)-\dots$$
But for any $P_{k}$; $|P_{k}|=f(n-2)$ therefore, $|P_{1}|=|P_{2}|=\dots=|P_{\frac{n}{2}}|$and in general, given any $k$ where $1\le k\le \frac{n}{2}$, then $|P_{1}\cap P_{2}\cap\dots\cap P_{k}|=f(n-2k)-\dots$
So the number of sets not containing any of the original pair is:
$$f(n)-\left(f(n-2)+f(n-2)+\dots\right)+\left(f(n-4)+f(n-4)+\dots\right)-\dots$$
which equals:
$$f(n)-{\frac{n}{2}\choose 1}f(n-2)+{\frac{n}{2}\choose 2}f(n-4)-\dots$$
So in the case where $n=8$
$$f(8)-4f(6)+6f(4)-4f(2)+1f(0)=105-60+18-4+1=60$$
Best Answer
I would say this (not very different from your justification): suppose you enumerate the $n$ people from a specific person, whatever its position on the ring. Two different arrangements then correspond exactly to two different permutations of the $n-1$ remaining people. Hence there are $(n-1)!$ such arrangements.