Obviously you can factor a real number into any set of numbers whose product is that real number. As for PRIME factorizations I know of one generalization of that probably DOESN'T work well for real numbers:
If you consider prime factorizations as finite sequences of exponents of the first primes (up until the sequence ends), then positive integers represent the special case where these exponents are all non-negative integers:
$$ 2^3 3^2 5^0 7^1 = 504 $$
If you allow negative integers, then you get unique prime factorizations for all positive rational numbers:
$$ 2^3 3^{-2} 5^0 7^1 = \frac {56} {9}$$
If you allow any rational number exponents (positive, zero, or negative), then you get all rational POWERS of positive rational numbers:
$$ 2^{\frac {3} {2}} 3^{- \frac {2} {3}} 5^1 = \left(\frac {2560} {81}\right)^{\frac {1} {6}}$$
You can keep going by expanding the set of exponents; however, as soon as your set of exponents includes numbers of the type $ \log_p q $, where p and q are different prime numbers, then you looselose uniqueness of prime factorizations:
$$ 2^{\log_2 {3}} = 2^0 3^1 = 3$$
so, if you just let any real number be an EXPONENT, you looselose uniqueness. Technically, since each positive real number has at least one prime factorization, you can make a rule to choose which one is the primary one, and make a set of exponent sequences that only includes those, though maybe you'd have to give each prime a different set of exponents it can accept and you might have to looselose closure under addition or multiplication or something. In the context where I wanted these, is was natural to expect all these sets of numbers I wanted prime factorizations for to be closed under multiplication, which meant the sets of exponent sequences had to be closed under element-wise addition.
Also, just to state it out clearly, you can consider these sequences of expnentexponent to infinite, but all zeros after a certain point, since there are infinite primes for them to be exponents of. You can also then view these as vectors in a countably infinite-dimensional space, if that makes the element-wise addition seem more natural, or if it helps you to see the conversion to the numbers they are prime factorizations of as using a dot product:
$$ r = b^{\vec M(r) \cdot [\log_b 2, log_b 3, log_b 5, ...]^T} $$
where $\vec M$ is the function that converts the number r to its "prime factorization", and b is any number that can be a base of a logarithm (I would be inclined to pick either b=2 or b=e as a default, or maybe $b=2^{1/12}$ for music theory or b=10 for some other kind of convenience, but it doesn't matter what you pick.) $\vec M$ is only a function if "prime factorizations" are unique of course.
However, if you allow infinite sequences of exponents that DON'T end in all zeros, you run into problems. Firstly, in the case where exponents have to be non-negative integers, any sequence with infinitely many non-zero exponents will lead to a product that diverges (so, in a sense, infinity clearly has no unique prime factorization). If you allow negative and/or rational exponents, then you can get products that actually converge, but you'll probably looselose uniqueness, and also the set of possible products will be larger than with finite sequences of exponents. (I wouldn't be surprised if you can basic Analysis axioms and theorems and logarithms to prove that you can reach any positive real number this way with just negative exponents, or any real number greater than or equal to one with just positive rational exponents.)
If you want negative numbers, you can introduce -1 as a "prime number", but it behaves weirdly with loss of uniqueness and introduction of complex numbers if you allow it to take the same exponents as the actual primes. You can also use $i = \sqrt {-1} $ instead of -1 and get similar issues. I don't know what happens if you use Gaussian primes or "primes" of other rings, but it would probably work better than with using units like i and -1.
