Can any meromorphic function be represented as a product of zeroes and poles?

Short answer: Not all, but...

Turns out this is more complicated than user170431's answer suggested, though it hinted at the correct direction.

Given a meromorphic function $f(z)$ with poles at $\{p_i\}$ of multiplicites $\{n_{p_i}\}$, one can indeed use Mittag-Leffler's theorem to obtain a holomorphic function

$$h(z) = f(z) - \sum_i \pi_{p_i}(1/(z-p_i))$$

where $\pi_{p_i}$ denotes a polynomial of degree $n_{p_i}$, and due to their finite order (meromorphic functions don't have essential singularities) one can expand $f(z)$ to the common denominator.

What the other answer and my question's assumption did however neglect was the full extend of the Weierstrass factorization theorem, according to which $h(z)$ actually factorizes into

$$h(z) = e^{g(z)}\prod_j(z-z_j)^{n_{z_j}}$$

where $n_{z_j}$ denotes the multiplicity of the zero $z_j$ and $g(z)$ is an entire function. Note that I simplified matters here, for infinitely many zeros (or of infinite multiplicity) parts of the exponential are inside the product as "elementary factors" in order to have the infinite product converge. In my sloppy notation, they are part of $\exp g(z)$ though.

Therefore, the actual representation of a meromorphic function is

$$f(z) = e^{g(z)}\prod_i(z-z_i)^{n_i}$$

(now the $z_i$ also include the poles again, as in the question's notation). Only if there are finitely many zeros of finite multiplicities (i.e. a quotient of two polynomials), the representation without an exponential in front (i.e. a constant $g(z)$) is valid.

Yes, this is Mittag-Leffler's theorem combined with the Weierstrass factorization theorem after reducing to the common denominator.