Prove that $G$ is an abelian group if $\{(g, g):g\in G\}$ is a normal subgroup.

I think I've got it!

Let $g=a$.

Then, $aa=g'a\implies g'=a$

So, $bg=g'b\implies ba=ab$

So, $G$ is abelian.

I think you've solved your own question correctly. Kudos. Here's a slightly different, perhaps more direct, approach:

$$D\lhd G\times G\iff\forall\,g\in G\;,\;\;(a,b)(g,g)(a,b)^{-1}\in D\iff$$

$$(aga^{-1}\,,\,bgb^{-1})\in D$$

and this means $\;aga^{-1}=bgb^{-1}\;$ for all $\;a,b,g\in\Bbb G\;$ . Take now $\;g=b\;$ , and the above says that for all $\;a,b\in G\;$ we have

$$aba^{-1}=bbb^{-1}=b\implies ab=ba\;\;\;\;\;\;\;\;\square$$

Since D is normal in GxG So using the property of cosets and for (g,g) in GxG We have => D(g,g)=(g,g)D . Let (d,d) in D ve arbitray where d is in G . =>(d,d)(g,g)=(g,g)(d,d) => (dg,dg)=(gd,gd) . Now comparing corresponding entries we have dg=gd and g,d both are in G . Hence G is abelian