Subgroups of $G\times H$

Is it true that if $$G,H$$ are groups, then every subgroup $$S$$ of $$G\times H$$ can be written in the form $$S\cong S_G\times S_H$$, where $$S_G$$ and $$S_H$$ are subgroups of $$G$$ and $$H$$ respectively? That is, is $$S$$ isomorphic (not necessarily equal) to a direct product of subgroups of $$G$$ and $$H$$?

Notice that I am not asking whether $$S=S_G\times S_H$$ for some $$S_G,S_H$$ subgroups of $$G,H$$; this is obviously false. Wikipedia provides a trivial counterexample: consider the subgroup of $$G\times G$$ consisting of elements in the form $$(g,g)$$. This is clearly not a direct product of subgroups of $$G$$. However, it is isomorphic to a direct product of subgroups of $$G$$; namely, the product $$G\times \{e\}$$.

If you set $$G = H = S_3$$, the symmetric group of order 6, then consider the subgroup $$K$$ of elements $$(x,y)$$ where either both $$x$$ and $$y$$ are even or both $$x$$ and $$y$$ are odd. It's not hard to see that $$K$$ is a subgroup of $$S_3 \times S_3$$ of order 18.
Now if $$K$$ was to have a decomposition $$K \cong G_1 \times G_2$$ where $$G_1, G_2$$ are isomorphic to subgroups of $$S_3$$, then by consideration of order one of $$|G_1|,|G_2|$$ would be equal to 6 and the other would be equal to 3. Without loss of generality set $$|G_1| = 6, |G_2| = 3$$. Let $$x$$ be an element of $$G_1$$ of order 2. Then $$(x,1)$$ must commute with all elements of $$G_1 \times G_2$$ of form $$(1,y)$$ for all $$y$$ in $$G_2$$. This includes the two non-identity elements of $$G_2$$ that are of order 3. This provides two commuting elements $$(x,1)$$ and $$(1,y)$$ of $$G_1 \times G_2$$ of orders 2,3, respectively.
However, it is straightforward to check that the elements of order 2 in $$K$$ do not commute with any of the elements of order 3 in $$K$$. The elements of $$K$$ of order two are all of the form $$(t_1, t_2)$$ where $$t_1$$ and $$t_2$$ are both transpositions. Furthermore, all nine such elements are conjugate (as any two conjugates in $$S_3$$ are conjugate), meaning that for any given element $$(t_1,t_2)$$, the centralizer is exactly given by $$\langle t_1 \rangle \times \langle t_2\rangle$$, which is a Klein four subgroup containing no elements of order 3.
Thus, by contradiction, $$K$$ does not decompose as a direct product of form $$G_1 \times G_2$$.