This is what happens if one cares not for the differerence between self-adjointness and Hermiticity, or, more general, if one ignores that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space.
Let's set $a=1$ for convenience. The operator $p =-\mathrm{i}\hbar\partial_x$ acting on wavefunctions with periodic boundary conditions defined on $D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land \psi'\in L^2([0,1])\}$ is Hermitian, but not self-adjoint, that is, on the domain of definition of $p$, we have $p=p^\dagger$, but $p^\dagger$ admits a larger domain of definition. The Hermiticity of $p$ follows from the periodic boundary conditions killing the surface terms that appear in the $L^2$ inner product $$\langle \phi,p\psi\rangle - \langle p^\dagger \phi,\psi\rangle = \int\overline{\phi(x)}\mathrm{i}\hbar\partial_x\psi(x) - \overline{\mathrm{i}\hbar\partial_x\phi(x)}\psi(x) = 0$$ for every $\psi\in D(p)$ and every $\phi\in D(p^\dagger) = \{\phi\in L^2([0,1])\mid \phi'\in L^2([0,1])\}$. Since $D(p)\subset D(p^\dagger)$ but $D(p)\neq D(p^\dagger)$, the operator p is Hermitian, but not self-adjoint, and hence not the correct observable for momentum.
In particular, the functions $\psi_{p_0}(x) = \exp(\mathrm{i}p_0x),p_0\in\mathbb{C}$ belong to $D(p^\dagger)$, but not to $D(p)$, so they are not eigenvectors of $p$ and thus not solutions to the Schrödinger equation. $p$ has no eigenvalues, but those of $p^\dagger$ are all complex numbers $p_0\in\mathbb{C}$.
The operator $p$ can be made-self-adjoint by enlarging its domain of definition. Recall that a phase is unphysical - this is the motivation to relax the boundary condition to have $$ D(p) := \{\psi\in L^2([0,1])\mid \exists\alpha\in\mathbb{R}:\psi(0) = \mathrm{e}^{\mathrm{i}\alpha}\psi(1)\land \psi'\in L^2([0,1])\}$$ and for this new operator, we have $p=p^\dagger$ and $D(p^\dagger) = D(p)$. Its spectrum is the real line, the eigenfunctions are $\psi_{p_0},p_o\in\mathbb{R}$.
Now, for the question of the commutator: the multplication operator $x$ is defined on the entire Hilbert space, since for $\psi\in L^2([0,1])$ $x\psi$ is also square-integrable. For the product of two operators $A,B$, we have the rule $$ D(AB) = \{\psi\in D(B)\mid B\psi\in D(A)\}$$ and $$ D(A+B) = D(A)\cap D(B)$$ so we obtain \begin{align} D(px) & = \{\psi\in L^2([0,1])\mid x\psi\in D(p)\} \\ D(xp) & = D(p) \end{align} and $x\psi\in D(p)$ means $0\cdot \psi(0) = \mathrm{e}^{\mathrm{i}\alpha}\psi(1)$, that is, $\psi(1) = 0$. Hence we have $$ D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1]) \land \psi(1) = 0\}$$ and finally $$ D([x,p]) = D(xp)\cap D(px) = \{\psi\in L^2([0,1])\mid \psi'\in L^2([0,1])\land \psi(0)=\psi(1) = 0\}$$ meaning the plane waves $\psi_{p_0}$ do not belong to the domain of definition of the commutator $[x,p]$ and you cannot apply the naive uncertainty principle to them. However, for self-adjoint operators $A,B$, you may rewrite the uncertainty principle as $$ \sigma_\psi(A)\sigma_\psi(B)\geq \frac{1}{2} \lvert \langle \psi,\mathrm{i}[A,B]\rangle\psi\rvert = \frac{1}{2}\lvert\mathrm{i}\left(\langle A\psi,B\psi\rangle - \langle B\psi,A\psi\rangle\right)\rvert$$ where the r.h.s. and l.h.s. are now both defined on $D(A)\cap D(B)$. Applying this version to the plane waves yields no contradiction.