Electromagnetic wave phasor conversion
I'm a couple months late ...

Your math checks out, but your words are imprecise.

When we describe a quantity in phasor notation, we drop the \$e^{+j\omega t}\$ time dependence. That's because the frequency and time dependence should be implied. (Note that typically electrical engineers use the \$e^{+j\omega t}\$ time dependence convention, while physicists often use \$e^{-j\omega t}\$.)

The relationship between the electric field in the time domain \$\mathcal{E}(x,y,z,t)\$, and the electric field phasor \$\mathbf{E}(x,y,z)\$ is

$$\mathcal{E}(x,y,z,t) \equiv \text{Re}\left\{ \mathbf{E}(x,y,z) e^{+j\omega t}\right\} $$

in your case, you should write

$$\mathbf{E}(x,y,z) = 10^{-4}\left( \mathbf{\hat{x}} e^{j(20z)} + \mathbf{\hat{y}}  e^{j(20z+\pi/2)} \right)$$
$$\mathcal{E}(x,y,z,t) = 10^{-4}\left( \mathbf{\hat{x}} \cos(\omega t+20z) - \mathbf{\hat{y}}  \sin(\omega t+20z) \right)  $$

Reflected vs transmitted, depends on the direction of an incident field, which you haven't told us. What I can tell you is that your wave is propagating in the \$-z\$ direction.

It is circularly polarized, but I don't know what you mean by "constant at all angles". It's circularly polarized because the polarization is rotating in a circle. At \$t=-\frac{1}{\omega}20z\$ the field is entirely polarized in the \$+\mathbf{\hat{x}}\$ direction. Increase the time to \$t=-\frac{1}{\omega}20z + \frac{1}{\omega}\frac{\pi}{2}\$, and it's polarized in the \$-\mathbf{\hat{y}}\$ direction. Increase in time another \$\frac{1}{\omega}\frac{\pi}{2}\$ and it's \$-\mathbf{\hat{x}}\$ polarized, another one and it's \$+\mathbf{\hat{y}}\$. Circular.

Also, if we want to be super picky, we should note that \$20\$ isn't unitless. It's \$20 \text{meters}^{-1}\$, or whatever length unit you're using for \$z\$.