Here is a proposition from Artin:
Usually when one wants to prove that $G\simeq H\times K$, one verifies the conditions of $(d)$. But if $(d)$ holds, then $G$ is isomorphic to $H\times K$ via the specific map $(h,k)\mapsto hk$. But theoretically it may happen that for $G$, the conditions in $(d)$ do not hold (and hence $G$ is not isomorphic to the product $H\times K$ via the product map), but $G$ is isomorphic to $H\times K$ via $(h,k)\mapsto h^{17}k^{-1}hkhk^2h^{-28}k$ or a similar map. Is my understanding correct? If so, what is an example when the above holds? And if $(d)$ does not hold, then how do I see whether there is an isomorphism between $G$ and $H\times K$ different from $(h,k)\mapsto hk$?
If you're not using the multiplication map from $H \times K$ to $G$, then $H$ and $K$ aren't really specific subgroups of $G$ anymore. They're just abstract groups that each happen to be isomorphic to some subgroup of $G$, but that identification has nothing to do with the isomorphism from $H \times K$ to $G$.
For example, let $G = \mathbb{Z} \times \mathbb{Z}$, and let $H$ and $K$ both be the first factor of $\mathbb{Z}$. Then the multiplication map isn't injective, so it's not an isomorphism, but abstractly $G$ is still isomorphic to $H \times K$.