The evidence seems to support that

Suppose the two numbers are $x,y$ . Now $x = y = 0$ satisfies the given conditions, but the ratio $\dfrac{0}{0}$ is undefined, i.e., not (definitively) $0$ , so the answer is $\boxed{\text{No}}$ .

Comments: Out of interest, we'll determine all possible pairs $x,y$ which satisfy the given three conditions. From the first condition we have that $x + y$ equals one of $x,y$ ; without loss of generality let $x + y = x \Longrightarrow y = 0$ .

From the third condition we have that $xy = 0$ , which will hold for any $x$ since $y = 0$ .

From the second condition we can have any of (i) $x - 0 = x$ , a tautology, (ii) $x - 0 = 0 \Longrightarrow x = 0$ , (iii) $0 - x = x \Longrightarrow x = 0$ or (iv) $0 - x = 0 \Longrightarrow x = 0$ .

So one of $x,y$ must be $0$ , but the other can be any real number. Now for $x \ne 0$ we could have the ratio being $\dfrac{0}{x} = 0$ , which is indeed one of the two numbers, but as this is not the case when $x = 0$ we must conclude that the statement in question is not true for all cases.