It's easy to see that $(ab, bc, ca, d)$ is a Pythagorean quadruple. In order to find all solutions we use the primitive Pythagorean quadruples. That is
$$ \begin{array} aab= m^2+n^2-p^2-q^2 \\ bc=2(mq+np) \\ ca=2(nq-mp) \\ d=m^2+n^2+p^2+q^2 \end{array}$$
Where $m$, $n$, $p$, and $q$ are non-negative integers with greatest common divisor $1$ such that $m + n + p + q$ is odd.
One can easily conclude that
$$ \begin{array} aa^2= \frac{(m^2+n^2-p^2-q^2)(nq-mp)}{mq+np} \\ b^2= \frac{(m^2+n^2-p^2-q^2)(mq+np)}{nq-mp} \\ c^2=\frac{(mq+np)(nq-mp)}{m^2+n^2-p^2-q^2} \end{array}$$
How can we find all quadruples of $(m, n, p, q)$ that make above relations true?
For the equation.
$$a^2b^2+c^2b^2+a^2c^2=d^2$$
The solution can be written directly.
$$b=t(tp^2+2kps+ts^2)$$
$$a=kt(p^2-s^2)$$
$$c=2kp(kp+ts)$$
$$d=kt((2k^2+t^2)p^4+6ktsp^3+2(k^2+t^2)p^2s^2+2ktps^3+t^2s^4)$$