Does $HK \cong H \times K$ imply that $H$ and $K$ satisfy the Internal Direct Product criteria? Let $H$ and $K$ be subgroups of a group $G$, and suppose that $HK \cong H \times K$. 

Does it then follow that that $HK$ is the internal direct product of $H$ and $K$? In other words, does it follow that both $H$ and $K$ are normal in $HK$, and that $H \cap K = \{e\}$?
Not necessarily - it depends on what the isomorphism is.

Say $G=\mathbb{Z}_2^{\omega}$ and take $H=\mathbb{Z}_2\times \mathbb{Z}_2 \times 0 \times 0 \times 0 \dots\;\; (\simeq \mathbb{Z}_2 \times \mathbb{Z}_2),$ $K=0 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \dots \;\;(\simeq \mathbb{Z}_2^{\omega})$.

Then $HK=G=\mathbb{Z}_2^{\omega}\simeq (\mathbb{Z}_2\times \mathbb{Z}_2) \times \mathbb{Z}_2^{\omega} \simeq H \times K$, but $H \cap K = 0 \times \mathbb{Z}_2 \times 0 \times 0 \dots \;\; \neq 0$.

However, assuming both $K,H$ are normal in $G$, you can say the following:


  Assume that the map $H \times K \rightarrow HK, \;\; (h,k) \mapsto hk$
  is an isomorphism. Then $H \cap K = \{e\}$.
Yes that's right since the intersection is the identity e. And  h and k are normal subgroup it will follow that h×k =hk