I think this is a combination of both a convention and a physical problem. You are equating the energy eigenvalue (ie, the total energy) to an expression that contains only $x_{ZPF}$, and does not contain $p$ at all. In other words, you are equating the total energy to a potential energy. This would be analogous to equating $E_\mathrm{total} = \frac{1}{2}kA^2$ to find the amplitude $A$ of a classical harmonic oscillator. The result is that you are using $x_{ZPF}$ to mean the "amplitude" of the zero-point fluctuation. The true result, as Ondrej Cernotik's answer derives, uses the rms value $x_{ZPF} = \sqrt{\langle\hat x^2\rangle}$. So that's the sense in which it is a convention.
The sense in which it is a real physical problem is that the "amplitude" of a quantum oscillator isn't really a well-defined, measurable thing. The quantum oscillator has a non-zero probability amplitude going all the way out to infinity. The rms value is well-defined and easy to measure. So that's the preferred definition.
I'll start with things you already understand. The question considers a model with Hamiltonian
$$
H \propto x^2+p^2-1
\tag{1}
$$
with a positive overall factor, where $x$ and $p$ are operators satisfying
$$
[x,p]=i.
\tag{2}
$$
As noted in the question, the Hamiltonian can also be written
$$
H\propto a^\dagger a
\tag{3}
$$
with
$$
a\propto x-ip.
\tag{4}
$$
The eigenstate of $H$ with the lowest eigenvalue is the state $|0\rangle$ that satisfies
$$
a|0\rangle = 0.
\tag{5}
$$
This is the ground state. The eigenvalue (which is sometimes called "zero point energy") is irrelevant, because shifting $H$ by a constant term does not change the fact that (5) has the lowest eigenvalue among all eigenstates of $H$. You already understand this.
I'm not sure what you want to call "kinetic energy" in this context, but suppose it's the operator
$$
K \propto p^2.
\tag{6}
$$
The ground state an eigenstate of the total energy operator $H$, so measuring $H$ in the ground state would always give the same outcome. But the ground state is not an eigenstate of the kinetic energy operator $K$. If we measure $K$ in the state $|0\rangle$, and if we repeat this experiment a jillion times, we'll get a jillion different outcomes. This is an example of the so-called uncertainty principle: if you measure an observable in a state that is not one of the observable's eigenstates, you get diverse outcomes. Sometimes people describe this using words like "vacuum fluctuations," but I don't know why. Maybe they're imagining Hidden Variables.
When we say that a system is at absolute zero, we mean that it is in its ground state — the state of lowest total energy. In this example, that's the state satisfying (5). In some systems, like an ideal gas, total energy and kinetic energy are the same thing. The harmonic oscillator is not one of those systems, at least not if "kinetic energy" means (6). In general, zero absolute temperature does not mean zero kinetic energy. Zero absolute temperature does mean time-independent, and the ground state — the ray represented by the vector $|0\rangle$ — is indeed independent of time.
Best Answer
Short answer: By the uncertainty principle, the harmonic oscillator can't be localized at the minimum value of potential energy, i.e., $x=0$, because, by the uncertainty principle, it's momentum would become large (strictly speaking, the expectation value of $p^2$, and thereby it's kinetic energy, becomes large). The lowest energy state of the harmonic oscillator is a compromise between minimizing potential energy (i.e., $x^2$) and kinetic energy (i.e., $p^2$), which cannot be done simultaneously, because $\langle x^2\rangle \langle p^2 \rangle \ge \frac{\hbar^2}{4}$.