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In particular, can a simple group (so that we don't have any proper normal subgroups) have an exact factorization, i.e. do there always exist $H, K

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    What do you mean by this in the case that $|G|$ has more than two prime factors? – 2017-02-23
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    @EricWofsey I think the question in the form stated in the body is fine, but I agree that the version in the title is not great. – 2017-02-23
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    But there are two separate questions here, so the poster should clarify which one is being asked! – 2017-02-23

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To answer the question in the title, the group ${\rm PSU}(3,3)$ of order $6048 = 2^5.3^3.7$ cannot be wriiten as a product $P_1P_2P_3$ of $3$ Sylow subgroups in any order. That is proved in a paper I wrote with Petey Rowley a while ago: D. Holt and P. Rowley, On products of Sylow subgroups in finite groups, Arch. Math. 60 (1993), 105-107.

The same group is a counterexample to the question in the body because you can check by computing all subgroups of ${\rm PSU}(3,3)$ on a computer that there are do not exist inters $a,b>1$ with $ab=6048$ such that this group has subgroups of orders $a$ and $b$. There may be smaller counterexamples.