$b$ cannot be zero because we can't divide by zero. $ab=\frac{a}{b}\Rightarrow b^2=1$ so $b=\pm 1$.

If $b=1$ then the equation $a+b=ab$ becomes $a+1=a$ which is impossible. Hence we must have $b=-1$. Then $a+b=ab$ becomes $a-1=-a\Rightarrow a=\frac{1}{2}$.

Finally we check that $\frac{a}{b}=a+b$ holds for $(a,b)=(\frac{1}{2},-1)$ which it does, so there is exactly one pair which satisfies the conditions.

*This problem is taken from the UKMT Mathematical Challenges.*