The boundary conditions, as noted in the comment below the original post, are:
\begin{align*}
E_{1z} &= E_{2z} & \epsilon_r E_{1y} &= E_{2y} \\
B_{1y} &= B_{2z} & B_{1z} &\approx B_{2z}
\end{align*}
(I'm using 1 to denote the dielectric and 2 to denote the air. The "approximate equals" sign above is because we're assuming $\mu_1 \approx \mu_2$.)
These are pretty easy to solve for $\vec{E}_2$ and $\vec{B}_2$, as noted above; the results are
$$
\vec{E}_2 = (5 \epsilon_r \hat{j} + 10 \hat{k}) \cos(\omega t - kx)
$$
$$
\vec{B}_2 = (10 \hat{j} - 5 \hat{k}) k \sin(\omega t - kx)
$$
These appear to violate Maxwell's equations, assuming that the fields do not depend on $y$ or $z$:
$$
\nabla \times \vec{E}_2 = - \frac{\partial E_{2z}}{\partial x} \hat{y} + \frac{\partial E_{2y}}{\partial x} \hat{z} = (10 \hat{j} - 5 \epsilon_r \hat{k} ) k \sin(\omega t - kx)
$$
$$
- \frac{\partial \vec{B}_2}{\partial t}= (10 \hat{j} - 5 \hat{k} ) \omega \sin(\omega t - kx)
$$
But what's important to note here is that these are only the field values at $y = 0$. In fact, this is just telling us that $\partial E_x/\partial y \neq 0$ along the interface.
(ETA: what's below this point is probably not a good way to think about things. See edit below.)
In fact, it might be possible to think of this situation as the limit of $\theta \to \pi/2$ of total internal reflection. Suppose you had a wave traveling in the $xy$-plane towards the interface in the diagram above, with its polarization in the plane of reflection. This would give rise to a reflected wave in the dielectric, and an evanescent wave in the air. This evanescent wave would, in general, have non-zero $E_x$ and $E_y$, and since everything dies off exponentially in the $y$-direction, we would have $\partial E_x/\partial y \neq 0$. I suspect (although I haven't proven it) that you can view your problem as the case of total internal reflection in the case of grazing incidence, and that the fact that $\partial E_x/\partial y \neq 0$ is just a manifestation of evanescent waves in the air in this limit.
What's less than satisfying about this explanation, of course, is that it demands that the electric field gains a component in the $x$-direction in the air even though it has no $x$-component in the dielectric. I honestly don't know enough about evanescent waves to know whether this is a dealbreaker for this interpretation or not.
EDIT: Having gone through the calculations, I'm not as sure about the evanescent-wave interpretation. The basic idea I had was that you could write down the usual three-wave solution at an interface (incident, transmitted, and reflected), use the Fresnel equations to find the total $E_x$ above and below the interface, and then show that if you took the limit appropriately as the incidence angle $\theta \to \pi/2$, you could get a situation where $E_x \to 0$ in the dielectric but $E_x \nrightarrow 0$ in the air. However, assuming I did the algebra correctly, the ratio of the $E_x$ values immediately above & below the interface will be
$$
\frac{\tilde{E}_{x,\text{dielec}}}{\tilde{E}_{x,\text{air}}} = \frac{i\sqrt{\epsilon \sin^2 \theta - 1} + \cos \theta}{\sqrt{\epsilon} \sqrt{\epsilon \sin^2 \theta - 1}},
$$
which approaches $i/\epsilon \neq 0$ as $\theta \to \pi/2$. Thus, you can't view this situation as the grazing-incidence limit of total internal reflection.
I'm still moderately confident in my answer above (that $\partial E_x /\partial y \neq 0$ at the boundary, even though $E_x$ itself vanishes), but so far I can't address the unsatisfactory aspects of this answer that I noted above.
