Greatest common divisor proof attempt

Cancelling $\,d = (a,b)\,$ from both sides reduces it to the case where $\,(a,b) = 1.\,\,$ Then it becomes $\, 2 = (a-b,\,a+b)\,$ for $a,b\,$ odd and coprime. Proof: $ $ clearly $\,2\mid a-b,a+b\,$ by $\,a,b\,$ odd. Conversely $\,c\mid a-b,a+b\,\Rightarrow\, c\mid 2a,2b\,\Rightarrow\, c\mid (2a,2b)=2(a,b)=2.\ $ QED

Alternatively: $\,a\!+\!b,\,a\!-\!b\,$ are divisible by the coprimes $\,2\,$ and $\,(a,b)\,$ so also by their product $2(a,b) = (2a,2b),\,$ so $\,(2a,2b)\mid (a\!+\!b,a\!-\!b).\,$ This completes the proof (with your converse).

Let $D=\gcd(2a,2b)$ and $d=\gcd(a+b,a-b)$

Clearly $2\,|\,D$ so let $D=2E$. As you note, $E=\gcd(a,b)$. It is clear that $2\,|\,d$, and it is clear that $E$ is odd. To see that $E\,|\,d$ we remark that $E$ must divide both $a$ and $b$, hence $E$ divides $a\pm b$ and we are done.

$ D=(a,b)$ so that $D|a,\ D|b,\ 2D=(2a,2b)$ and $D$ is odd

Hence $D|a+b,\ D|a-b$ so that $D|(a+b,a-b)$

Since $a+b,\ a-b$ are even, $2D|(a+b,a-b)$

When $D'=2kD|(a+b,a-b),\ k>1$, then $$2kD=D'|(a+b)-(a-b)=2b ,\ kD|b$$

Similarly, $kD|a$ so that it is a contradiction to $D=(a,b)$