Is there a "canonical" representation of integers using numbers other than primes? Consider the (cumbersome) statement: "Every integer greater than 1 can be written as a unique product of integers belonging to a certain subset, $S$ of integers.
When $S$ is the set of primes, this is the Fundamental Theorem of Arithmetic. My question is this: Are there any other types of numbers, for which this is true. 
EDIT: As the answers show, this obviously cannot be done. What if we relax the integer condition, i.e. can there be any other canonical representation of positive integers using complex numbers?
 A: If you mean that every positive integer gets a unique multiplicative factorization, then no, there is no other canonical representation. Why? Because then every prime number $p$ can be factorized, but the only way that's possible is if the components of the factorizations include the primes them-selves. Furthermore, you can't add any other number to the list because then the factorization of this number would be non-unique.
Alternatively, there are non-multiplicative representations of integers. The $p$-adic representation is just writing $n$ in "base $p$": $n=a_0+a_1p+a_2p^2+\cdots+a_rp^r$. Even though the golden ratio is not a rational number, we can write integers in base golden ratio.

Algebraic number theory studies number fields and rings of integers beyond just $\mathbb{Q}$ and $\mathbb{Z}$. Of note, there is not necessarily unique factorization of the elements. For example, in $\mathbb{Z}[\sqrt{-5}]$, we have
$$6=2\cdot3=(1+\sqrt{-5})(1-\sqrt{-5}).$$
This lead to some headaches (I assume anyway), until mathematicians figured out that even though the numbers don't factor uniquely, the ideals of the integers factor uniquely into products of prime ideals, which has lead to other algebraic constructions based off of them designed ultimately to study the structure of numbers. (If you don't understand this section of my answer, don't worry about it. It's for a later time then.)
A: No. Such a set $S$ must include the primes (because they have no other factors). If $s \in S$ is not prime, then it can be written as a product of primes, i.e., as a product of other elements of $S$. This contradicts uniqueness. Therefore the only such set $S$ is the set of the primes.
A: When you relax the integrality condition you start to pick up other solutions - for instance, the set of square roots of primes, $S=\{\sqrt{p}:p\in P\}$, is such that every number $n$ has a unique factorization $n=s_1^{e_1}s_2^{e_2}\ldots$ in terms of the elements of $S$; just consider the breakdown $n=p_1^{d_1}p_2^{d_2}\ldots$ and then set $e_1=2d_1$, etc.  On the other hand, this gives up existence : it's no longer the case that every sequence $e_1, e_2, \ldots$ of whole numbers corresponds to a unique integer; instead we have the additional requirement that all the $e_i$ be even.
More generally, any breakdown of primes into products of (non-integral) real factors will give you a solution: if $p_1 = t_{11}\cdot t_{12} \cdot \ldots\cdot t_{1m}$, $p_2 = t_{21}\cdot t_{22}\cdot\ldots\cdot t_{2n}$, and so on, then obviously any factorization of a number $n$ into products of $p_i$ extends to a (unique) factorization of $n$ into products of the $t_{ij}$.  What's more, any unique factorization of all numbers into real 'factors' $x_i$ has to come from this kind of breakdown, because obviously the prime numbers all need unique factorizations, and the factorization of a number $n$ into primes then 'extends through' the factorization of those primes into a factorization of $n$ in terms of the $x_i$, which by definition must be the factorization of $n$.  So you don't get anything particularly interesting by removing the non-integrality condition, just 'refinements' of the factorizations you already had.
