Decomposition of numbers: Show that there are positive integers $x, y$ and $z$, with $x$ and $y$ coprime, such that $a+b=x^2z\,$ and $\,c+d=y^2z$

You have the right idea with using the $\gcd$, but I don't know how to appropriately finish what you've done using the $\operatorname{lcm}$ as well. Instead, let

$$z = \gcd(a + b, c + d), \; \; a + b = ez, \; \; c + d = fz, \; \; \gcd(e, f) = 1 \tag{1}\label{eq1A}$$

For simpler algebra, let $j = ad - bc$. Then substitute \eqref{eq1A} into the equation that you gave to be satisfied to get

$$j^2 = (ez)(fz) = z^2(ef) \tag{2}\label{eq2A}$$

Thus, $z \mid j \; \; \to \; \; j = kz$, which substituting into \eqref{eq2A} and dividing by $z^2$ gives

$$k^2 = ef \tag{3}\label{eq3A}$$

As proven in multiple posts here, such as Show if a product of coprime numbers is a perfect square, so are the numbers - without FTA, a product of coprime integers being a square mean each of those integers is a square themselves. Thus, since \eqref{eq1A} gives that $\gcd(e, f) = 1$, then \eqref{eq3A} means that for some positive integers $x$ and $y$, coprime to each other, we have

$$e = x^2, \; \; f = y^2 \tag{4}\label{eq4A}$$

Using \eqref{eq1A}, this gives what you were asked to prove, i.e., $a + b = x^{2}z$ and $c + d = y^{2}z$.