algebra precalculus - Why the equation $3cdot0=0$ needs to be proven

In Algebra by Gelfand Page 21 ( for anyone owning the book).
He tries to prove that: $3\cdot(-5) + 15 = 0$.
Here's his proof: $3\cdot(-5) + 15 = 3\cdot(-5) + 3\cdot5 = 3\cdot(-5+5) = 3\cdot0 = 0$. After that he said:

The careful reader will asky why $3\cdot0 = 0$.

Why does this equation need to be proven ?
I asked somewhere and was told that $a\cdot0=0$ is an axiom which maybe Gelfand didn't assume was true during his proof.
But why does it need to be an axiom, it's provable:
In the second step of his proof he converted 15 to $3\cdot5$ so multiplication was defined so
$a\cdot0 = (0 + 0 + \cdots)$ x times $= 0$.
I'm aware multiplication is defined as repeated addition only for integers,
but 3 is an integer so this definition works in my example.

In case my question wasn't clear it can be summed up as:
Why he takes $3\cdot5=15$ for granted but thinks $3\cdot0=0$ needs an explanation?

Answer

Gelfand doesn't really take $3 \cdot 5 = 15$ for granted; in the ordinary course of events, this would need just as much proof as $3 \cdot 0$.

But the specific value $15$ isn't important here; we're really trying to prove that if $3 \cdot 5 = 15$, then $3 \cdot (-5) = -15$. That is, we want to know that making one of the factors negative makes the result negative. If you think of this proof as a proof that $3 \cdot (-5) = -(3 \cdot 5)$, then there's no missing step.

The entire proof could be turned into a general proof that $x \cdot (-y) = -(x\cdot y)$ with no changes; I suspect that the authors felt that this would be more intimidating than using concrete numbers.

If we really cared about the specific value of $3 \cdot 5$, we would need proof of it. But to prove that $3 \cdot 5 = 15$, we need to ask: how are $3$, $5$, and $15$ defined to begin with? Probably as $1+1+1$, $1+1+1+1+1$, and $\underbrace{1+1+\dots+1}_{\text{15 times}}$, respectively, in which case we need the distributive law to prove that $3 \cdot 5 = 15$. Usually, we don't bother, because usually we don't prove every single bit of our claims directly from the axioms of arithmetic.

Finally, we don't usually make $x \cdot 0 = 0$ an axiom. For integers, if we define multiplication as repeated addition, we could prove it as you suggest. But more generally, we can derive it from the property that $x + 0 = x$ (which is usually taken as a definition of what $0$ is) and the other laws of multiplication and addition given in this part of the textbook.