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
Examples and counterexamples.
Consider the following statements:
These two statements are clearly equivalent. This is an example of $$\forall x(P(x) \wedge Q(x)) \equiv \forall x P(x) \wedge \forall x Q(x)$$ where here the quantifiers range over my fish, $P(x) \equiv$ "$x$ is ugly" and $Q(x) \equiv$ "$x$ blows bubbles".
Now consider the following two statements:
So long as not all of my fish are goldfish, and not all of my goldfish are koi, the first statement will be true while the second will be false. This is an example of $$\forall x(P(x) \vee Q(x)) \not \equiv \forall x P(x) \vee \forall x Q(x)$$ where the quantifiers range over my fish, $P(x) \equiv$ '$x$ is a goldfish' and $Q(x) \equiv$ '$x$ is a koi'.
There are similar statements that illustrate the analogous (in)equivalences for $\exists$.
Proofs.
A proof that $\forall$ distributes over $\wedge$ is as follows:
$$\begin{align} \forall x(P(x) \wedge Q(x)) & \Leftrightarrow P(x) \wedge Q(x) & (\forall\text{-elimination})\\ & \Leftrightarrow P(x)\ \text{and}\ Q(x) & (\wedge\text{-elimination})\\ & \Leftrightarrow \forall xP(x)\ \text{and}\ \forall xQ(x) & (\forall\text{-introduction})\\ & \Leftrightarrow \forall xP(x) \wedge \forall xQ(x) & (\wedge\text{-introduction}) \end{align}$$
Sometimes $\forall$-elimination is called 'generalisation' or something similar... your notation and terminology may differ, but the steps in the proof should be roughly the same.
There is a similar proof for $\exists$.
Abstract nonsense.
The category theorist in me wants to give another proof: quantifiers are really functors (between preorder categories of formulae with certain free variables, ordered by entailment) which satisfy a string of adjunctions $\exists \dashv {*} \dashv \forall$. Now 'or' is a coproduct and 'and' is a product, and since right-adjoints preserve limits (and left-adjoints preserve colimits), we must have that $\forall$ distributes over $\wedge$ and $\exists$ distributes over $\vee$.