If $d$ is the greatest common divisor of all the numbers in set $B$, let $A = \{b/d : b \in B\}$. Then for each $a, b \in A$ ($a > b$) we have
$$\frac{a-b}{d(a,b)} \in A. \quad (*)$$
Observe that the greatest common divisor of the set $A$ equals $1$, therefore we can find a finite subset $A_1 \subset A$ for which $\gcd A_1 = 1$. We may think that the sum of elements of $A_1$ is minimal possible. Choose numbers $a, b \in A_1$ ($a > b$) and replace $a$ in the set $A_1$ with $\frac{a-b}{d(a,b)}$. The greatest common divisor of the obtained set equals $1$. But the sum of numbers decreases by this operation, which contradicts the minimality of $A_1$.

Thus, $A_1 = \{1\}$. Therefore all the numbers in the set $A$ have residue $1$ modulo $d$. Take an arbitrary $a = kd + 1 \in A$ and $b = 1$. Then $k \in A$ by $(*)$ and hence $k = ds + 1$. But $(k, kd + 1) = 1$, therefore $\frac{kd+1-ds-1}{d} = k - s = (d-1)s + 1 \in A$, so $s$ is divisible by $d$. But $s \in A$, therefore $s-1$ is also divisible by $d$, hence $d = 1$ (that means that $B = A$). Thus we have checked that if $a = kd + 1 = k + 1 \in A$ then $a-1 = k \in A$. Then all non-negative integers belong to $A$ because it is infinite.