The Univariate SoS Case
I gave a proof that any positive univariate polynomial is indeed SOS. But it seems my wits were wool-gathering, Alex correctly pointed out that I did not correctly account for the non-zero roots. Let me give the proof, more precisely this time.
Consider the roots of $P(x)$. Some are real, others are complex. The complex ones occur in conjugate pairs, so when you take the terms in the factorization $(x-a)(x-\bar{a})$ corresponding to them, you get $(x^2 + |a|^2)$.
For the real roots, the observation is that you cannot have a simple root: they must all be repeated. Indeed, if there was a simple root, the polynomial would dip below zero, and hence not be positive. In fact, the multiplicity of these roots must be even for the same reason. So the terms for these roots give us $(x-b)^2$-like terms.
Hence the factorization of $P(x)$ looks like:
The first set of terms (for the real roots) are square polynomials anyways, and multiplying out the second set of terms gives us an SoS polynomial. QED.
More on SOS
There is a fair bit of literature on the SoS hierarchy. The course by Barak and Kothari and Parillo and Steurer is a very good resource. Also this one by Bansal and Dadush. Or this one by Parillo has a different, more algebraic-geometry perspective. Another good CS-y intro to the Lasserre hierarchy, which directly deals with the SDP constraints, instead of the SoS proofs, is this one by Thomas Rothvoss (and here are his slides).