Does collision resistance stay when extending a hash function to a set domain? Given a Cryptographic hash function $h$ for element $x$, let's extend it to sets via $H(S)=\prod_{x\in{S}}{h(x)}$. I am asking if the new hash $H$ (in domain of set) is still collision resistant? 

To be more specific, is it the same difficult to compute $S$ for given $H(S)$, comparing to that of computing $x$ from given $h(x)$? 

Also how hard it is to find another $S'$ so that $H(S')=H(S)$? Is it at the same level of difficulty comparing to finding another $x'$ s.t. $h(x)=h(x')$.
If the hash outputs 160-bit integers, and the product operation is integer multiplication, then this is probably not secure, for reasons fgrieu outlines.  For instance, about $1/2^{19}$ of all 160-bit integers are $B$-smooth, if we take $B$ to be the $2^{20}$th prime.  Consequently, after hashing $2^{45}$ different candidate messages, we expect to find about $2^{26}$ smooth hash digests, which I suspect is enough to find a collision; and this computation seems likely to be feasible today.

On the other hand, if you work in a multiplicative group where the discrete log problem is hard, such as $\mathbb{Z}_p^*$, and if the hash function outputs random elements of this multiplicative group, then your scheme is secure.  For instance, if you pick a 2048-bit prime $p$, arrange for your hash function to output numbers that are uniformly distributed between $1$ and $p-1$, and interpret the product as "multiplication modulo $p$", then your scheme should be secure.  However, it might be a bit slow.  See the following papers for technical details and security analysis:


Incremental Multiset Hash Functions and Their Application to Memory Integrity Checking.  Dwaine Clarke, Srinivas Devadas, Marten van Dijk, Blaise Gassend, G. Edward Suh.  Asiacrypt 2003.  See the scheme they call MSet-Mu-Hash.
A new paradigm for collision-free hashing: Incrementality at reduced cost. Mihir Bellare, Daniele Micciancio.  Eurocrypt 1997.  See the scheme they call MuHash.