A simpler approach: $\gcd(a,b)=(a,b)=1$ if and only if there are integers $j,k\in\Bbb Z$ with $aj+bk=1$. Thus, since $(a+b,a-b)=1$, there are integers $j,k\in\Bbb Z$ so that $(a+b)j+(a-b)k=1$, but this is the same as $(j+k)a+(j-k)b=1$. Since $j+k,j-k\in\Bbb Z$, we have that $(a,b)=1$.
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 $c + d = y^{2}z$.
Best Answer
Suppose that $ab$ and $c$ were not coprime. Then there must be some prime number $p$ that divides both $ab$ and $c$. (Be sure to understand why).
But since $p$ is prime and $p\mid ab$ that implies that $p\mid a$ or $p\mid b$.
In the first case, we have then $p\mid a$ and $p\mid c$ contradicting that $\gcd(a,c)=1$, and in the second case we have $p\mid b$ and $p\mid c$ contradicting that $\gcd(b,c)=1$.