 Solution

An extension of $A$ by $B$ is a short exact sequence $1\to A\to G\to B\to 1$. It splits if the quotient map has a homomorphic section. Equivalently the embedded $A$ has a subgroup complement isomorphic to $B$, and $G$ is the corresponding semidirect product.

For $t\notin K$ and $a\in K$, both $t$ and $ta$ lie outside $K$, so $t^2=(ta{)}^2=1$. It follows that $tat=a^{-1}$, and $t=t^{-1}$ gives the claimed conjugation identity. Conjugation is an automorphism of $K$, so inversion is an automorphism there. Comparing the image of $ab$ with the product of the images gives $b^{-1}{a}^{-1}=a^{-1}{b}^{-1}$. Therefore $K$ is abelian.

Conversely inversion is an automorphism of any abelian group $A$, of order dividing two. Form the semidirect product $G=A⋊C_2$ with multiplicationFor completeness associativity follows because $b\mapsto b^{(-1{)}^{\epsilon }}$ is an automorphism and these automorphisms compose according to addition modulo two; $(1,0)$ is the identity and inverses follow from this formula. The subgroup $H=A\times \{0\}$ is normal with quotient $C_2$, and every outside element satisfies $(a,1{)}^2=(a{a}^{-1},0)=1$. Thus the generalized dihedral group supplies the required extension, including when $A$ is infinite or has elements of order two.