Pairs of integer pairs with same lcm, gcd and mean

Given $\operatorname{lcm}(x,y)\cdot\operatorname{gcd}(x,y)=|xy|$, we have $\operatorname{lcm}(a,b)\cdot\operatorname{gcd}(a,b)=\operatorname{lcm}(c,d)\cdot\operatorname{gcd}(c,d)$, so $|ab|=|cd|$. Given $\frac{a+b}{2}=\frac{c+d}{2}$, we have $c=(a+b)-d$. When substituted into $|ab|=|cd|$, this gives the two quadratics $d^2-(a+b)d\pm ab=0$. These have the two pairs of solutions $d=a,b$ (when $c,d$ are trivially a permutation of $a,b$) and $d=\frac{(a+b)\pm\sqrt{\Delta}}{2}$, where $\Delta$ is the discriminant $a^2+6ab+b^2$.

In the nontrivial case, $d$ is integral if $(a+b)\pm\sqrt{\Delta}$ is integral (and even), therefore $\Delta$ is a square. Hence, for some integer $k$, the triple $(a,b,k)$ is a solution of the 3-variable, 2nd-degree Diophantine equation $a^2+6ab+b^2=k^2$. This can be solved with a method analogous to finding Pythagorean triples: consider the intersections of the hyperbola $a^2+6ab+b^2=1$ with the line $a=m(b-1)$. The first intersection is $(1,0)$ and the second intersection is guaranteed to be rational when $m$ is integral, which paves the way for integer solutions when clearing denominators. The second solution is $b=\frac{(m+1)(m-1)}{m^2+6m+1},a=\frac{-2m(3m+1)}{m^2+6m+1}$, which we can substitute into $a^2+6ab+b^2=1$ and multiply through by $(m^2+6m+1)^2$, to give solutions for $(a,b,k)$, and by extension $(a,b,c,d)$ when substituted into the earlier equations. So the set of nontrivial solutions up to multiples and permutations is, for $m\in\mathbb{Z}$

$a=-2m(3m+1)\\ b=(m+1)(m-1)\\ c=-(3m+1)(m+1)\\ d=2m(1-m)$

For example, $m=-5$ gives the solution $(-140,24,-56,-60)$, which is a multiple and permutation of the solution $(-6,35,14,15)$ mentioned by the asker. This encompasses all solutions.