Two statements are equivalent if
$$A_1\Rightarrow A_2\text{ and } A_2\Rightarrow A_1.$$
More than two statements are equivalents if any two of them are equivalent.
So if you have for instance
$$A_1\Rightarrow A_2\Rightarrow A_3\Rightarrow A_1$$
then
$$A_2\Rightarrow A_1$$
so $A_1\iff A_2$. In the same way you could show that $A_2\iff A_3$ and $A_1\iff A_3$.
It is the basic idea, and it is working for any chain of length $n$ (you can prove it by induction).
UPDATE To add more context, the statement we were trying to prove is (see here)
Please note that neither your original Question nor your Addendum-Update is an accurate translation of the linked exercise. Here, I shall address your Addendum-Update (which really is a New Question)—which may have only a tenuous link to the motivating linked exercise.
Either $xy=y$ for all $x,y\in S$ or $xy=x$ for all $x,y\in S$
Here is the translation: $$\forall\, x,y\in S\:\,P(x,y) \;\;\lor\;\; \forall\, w,z\in S\:\,Q(w,z).\tag1$$
Which seemed very different to
For all $x,y\in S$ either $xy = y$ or $xy = x$.
Here is the translation: $$\forall\, x,y\in S\,\bigg(P(x,y)\lor Q(x,y)\bigg).\tag2$$
You are correct that $$(1)\not\equiv(2).$$
The latter is of the form $$\forall x P(x) \vee Q(x)$$
No it is not. Incidentally, please understand that $$\bigg(\forall x P(x)\bigg) \vee Q(x)\quad\text{ and }\quad\forall x\bigg( P(x) \vee Q(x)\bigg)$$ have different meanings, and that your suggestion is the same as the former, not the latter. In any case, none of these three is equivalent to the correct translation $(2).$
My friend's negation of the first statement is
There exist $x,y,z,w \in S$ such that $xy \neq x$ and $zw \neq w,$
while I think it should be
There exist $x,y\in S$ such that $xy \neq x$ and there exist $w,z\in S$ such that $wz \neq z$.
Which one is correct? Are they both wrong? Since it seems they are
most likely equivalent, I think they are both correct.
For ease of reading, here again is the first statement: $$\forall\, x,y\in S\:\,P(x,y) \;\;\lor\;\; \forall\, w,z\in S\:\,Q(w,z).\tag1$$
And here is its negation: $$\exists\, x,y\in S\:\,\lnot P(x,y) \;\;\land\;\; \exists\, w,z\in S\:\,\lnot Q(w,z)\tag{N1}$$
$$\exists\, x,y\in S\:\,xy\neq x \;\;\land\;\; \exists\, w,z\in S\:\,wz\neq z\tag{N1}$$
$$\exists\, x,y\in S\:\, \bigg(xy\neq x \;\;\land\;\; \exists\, w,z\in S\:\,wz\neq z \bigg)\tag{N1}$$
$$\exists\, x,y,w,z\in S\,\bigg(xy\neq x \land wz\neq z\bigg)\tag{N1}$$
Since your negation attempt is exactly the second line, while your friend's is exactly the fourth line, you are both correct!
Best Answer
As Brian M. Scott explains, they are logically equivalent.
However, in principle, the expression $$(*) \qquad A \Leftrightarrow B \Leftrightarrow C$$ is ambiguous. It could mean either of the following.
$(A \Leftrightarrow B) \wedge (B \Leftrightarrow C)$
$(A \Leftrightarrow B) \Leftrightarrow C$
These are not equivalent; in particular, (1) means that each of $A,B$ and $C$ have the same truthvalue, whereas (2) means that either precisely $1$ of them is true, or else all $3$ of them are true. Also, you can check for yourself that, perhaps surprisingly, the $\Leftrightarrow$ operation actually associative! That is, the following are equivalent:
In practice, however, (1) is almost always the intended meaning.