Basically, this is an equivalence relation on $B=A^4\setminus \{(a,b,c,d)\mid c+d=0\}$ where $(a_1,b_1,c_1,d_1)\sim (a_2,b_2,c_2,d_2)$ if $(c_1+d_1)(a_2+b_2)=(c_2+d_2)(a_1+b_1).$ You probably want to figure out an (estimate for) the size for $B$ and the maximum size of an equivalence class.

What if $A = \{0\}$? Will we have $B = \{\}$? What about $A = \{0,1\}$ and $B = \{0,\frac{1}{2},1,2\}$?

Is the condition maybe that $A$ is a set of positive real numbers?

@ThomasAndrews No, by no means, any finite subset of $\Bbb{R}$

Is there a reason you think this is true, given the counter-examples by vtand?

@VTand Good question. However, I simply found this problem without any more details than what I gave you. I assume that only division by $0$ is illegal for $B$ and the elements may not be distinct

@ThomasAndrews I rather tend towards thinking this question is poorly stated yet an interesting concept lies behind it

Since you haven't told us why you believe it is true, nor dealt with the counterexamples, it is worse than a poorly stated question. And you can fix your question.

Via approach0, I found that China Team Selection Test 2014 TST 1 Day 1 Q2 (AoPS thread) asks the same question, with the added restriction "$A$ is a finite set of positive numbers".

The only reason I believe it is true is because I have seen it on a math olympiad. And I answered to VTand's comment

It seems like you'd first want to find a lower bound on $C=A+A=\{a+b\mid a,b\in A\}.$ Then find the minimum number, given $C,$ of quotient of two elements of $C$ in terms of $|C|.$

So, you don't know what the question is, but want us to help you answer it? As written, the question is false.

@ThomasAndrews I made an edit with an additional restriction

The $2M-1$ part of the formula indicates to me positive values, since if $M$ is the set of possible values $\geq 1,$ then $2M-1$ would be the number of positive quotients.

The edit doesn't invalidate the counterexamples, unless what you meant is that no $a,b\in A$ has $a+b=0.$

@ThomasAndrews what about now?

@Sgg8 Your new restrictions do not make the question true. Try checking with $A = \{1, 2, 3, 4\}$. There are only $6$ possible fractions in $B$: $\frac{1+2}{3+4}$, $\frac{1+3}{2+4}$, $\frac{1+4}{2+3}$, $\frac{2+3}{1+4}$, $\frac{2+4}{1+3}$, $\frac{3+4}{1+2}$. Well, actually only $5$ distinct fractions, because $\frac{1+4}{2+3} = \frac{2+3}{1+4} = 1$.

Are you now just guessing? Edit your question so that it makes clear in the question - don't make a person read t9 the end and have the question mean something else. Rewrite your set as $\{(a+b)/(c+d)\mid a,b,c,d\in A\land c+d\neq0\},$ for example. Readers don't need to know the history of your question, just the question.

And distinct elements make $B$ smaller.

I edited the question as I still think the problem with positive reals deserves attention and may be useful to someone

Or should I close this question and make another one?

There is a handful of useful comments there already, so I tend towards editing the existing one being a better choice

If $A=\{1,2,\dots,n\}$ then, under the definition given originally, $$|B|=|\{(a,b)\mid 1\leq a,b\leq 2n,\gcd(a,b)=1\}|-2n,$$ where the $-2n$ comes from excluding $a=1,b>n$ and $b=1,a>n.$ This is somewhat famously approximated in size by $\frac{6}{\pi^2}(2n)^2-2n=\frac{24}{\pi^2}n^2-2n.$

I only mention that example because it seems like a worst case.

With the edit, this means you want $B'=\{x\in B\mid x\leq 1\}$ to have $|B'|\geq |A|^2.$

@ThomasAndrews i don't understand. After my edit this is literally the problem VTand gave a link to

Where did I say I had a pepblem with the edit? I was adding a suggesting for how to rephrase it so it might be easier to solve.

@ThomasAndrews well, now I do understand. Thank you for that