a+b=a*b, a-b=? if a and b are positive integers

Since $a =b(a-1),\;b|a$, and since $b=a(b-1),\;a|b$.

Therefore $a=b$, so $2a=a^2\implies a=2=b$.

$$a+b=ab$$ $$\frac{a+b}{ab}=1$$ This approach prevents either $a$ or $b$ being $0$. $$\frac1a+\frac1b=1$$ The only integer values of $a,b$ that satisfy this are $a=b=2$ ($a\gt2\implies1\lt b\lt2$).

We can see $a=b$ more directly by letting $b=ka$. Then $a(k+1)=ka^2\to a(k+1-ka)=0$. So either $a=0$ (and so $a=b$) or $a=\frac{k+1}{k}$. As $a\in\mathbb{Z}, k=1\to a=b=2$.

If we drop the constraint that $a\in\mathbb{Z}$, then for example consider $k=2$. Then $a=\frac32, b=3, a+b=ab=\frac92$.

Finally $a-b=ab-2b=b(a-2)$ which is only zero if $b=0$ or $a=2$.

$a+b = ab\\ a(1-b) +b = 0\\ a = \frac {b}{b-1}$

$b-1$ divides $b$ when $b = 0$ or $b-1 = 1$ i.e. everything divides $0,$ and $1$ divides everything.

$\gcd (b,b-1) = 1$ which can be demonstrated by the Euclidean algorithm.

There is no $b>2$ such that $(b-1)$ divides $b.$