From the given equations we have $a = c - b$, $d = 2c - a = b + c$. So
$$abcd = (c-b) \cdot b \cdot c \cdot (b+c) = bc(c^2-b^2) = \frac{1}{2}(2bc)(c^2-b^2).$$
Therefore $abcd$ is the area of a right triangle with legs $2bc$ and $c^2 - b^2$.
The hypotenuse of that triangle equals
$$\sqrt{(2bc)^2 + (c^2 - b^2)^2} = \sqrt{c^4 + b^4 + 2b^2c^2} = b^2 + c^2.$$
As $b$ and $c$ are positive integers and $c > b$ (since $c - b = a > 0$), the lengths of all three sides, $2bc$, $c^2 - b^2$ and $b^2 + c^2$ are positive integers.