I am trying here to answer your comment "I guess, if fields are undergoing a wave-like change in one frame of reference, they must undergo the same kind of change in any other frame of reference."
It's a legitimate remark and I would formulate it even more clearly: in a frame of reference where the charge is at rest it doesn't emit e.m. waves. But it can't be possible that in one frame we have an e.m. wave, and in another one we haven't.
Now, let me show you that although the e.m. field produced by the moving charge is time-dependent, it isn't similar to the e.m. waves which travel with the light velocity. For avoiding solving the equations in my previous answer, I'll use another procedure here. I will consider two frames of coordinate. A frame $O$ where the charge is at rest and in the origin of coordinates, s.t. we have only the Coulomb field,
$$\vec E = \frac {Q}{4\pi\epsilon_0} \frac {\vec r}{r^3}, \ \ \text {and} \ \ \vec B = 0, \tag{i}$$
and a frame $O'$ moving with a velocity $v$ in the direction $-z$. That means, in the frame $O'$ the charge moves with the velocity $v$.
Now, we can apply the special relativity transformations for the space and time
$$\begin{cases}
t = \gamma \left(t' + \frac {\vec r' \cdot \vec v}{c^2}\right) \\
\vec {r} = r'_{\perp} + \gamma \left(r'_{\|} + \vec v t' \right).
\end{cases} \tag{ii}$$
For the electric and magnetic field we have, by taking in consideration that in $O$ there is no magnetic field,
$$\begin{cases}
\vec {E'} = \gamma \vec E + (1 - \gamma) \frac {\vec E \cdot \vec v}{v^2} \vec v \\
\vec {B'} = -\gamma \frac {\vec v \times \vec E}{c^2}.
\end{cases} \tag{iii}$$
Let's introduce $\text {(i)}$ in $\text {(iii)}$.
$$\begin{cases}
\vec {E'} = \gamma \frac {Q}{4\pi\epsilon_0} \left[\frac {\vec r}{r^3} + (\gamma ^{-1} - 1) \frac {r_{\|}}{r^3} \frac {\vec v}{v}\right] \\
\vec {B'} = -\gamma \frac {Q \mu_0}{4\pi} \ \vec v \times \frac {\vec r}{r^3}.
\end{cases} \tag{iv}$$
The final thing is to replace $\vec r$ by the expression in $\text {(ii)}$,
$$\begin{cases}
\vec {E'} = \gamma \frac {Q}{4\pi\epsilon_0} \frac {\vec {r'_{\perp}} + \left[\gamma + (1 - \gamma)\frac {\vec v}{v}\right]\left(\vec {r'_{\|}} + \vec v t' \right)}{\left[ r'^2_{\perp} + \gamma^2 \left(r'_{\|} + v t' \right)^2\right]^{3/2}} \\
\vec {B'} = -\gamma \frac {Q \mu_0}{4\pi} \ \frac {\vec v \times \vec {r'_{\perp}}}{\left[r'^2_{\perp} + \gamma^2 \left(r'_{\|} + v t' \right)^2\right]^{3/2}}.
\end{cases} \tag{v}$$
As one can see, these are not travelling waves of type ~ $e^{i(kr - \omega t)}$ with $\omega = ck$. The formulas $\text {(v)}$ represent just a Coulomb field, that in the frame $O'$ appears as dragged after the charge $Q$, however, when expressing the position dependence of the field in terms of the coordinates of the frame $O'$ according to the special relativity, dependence of time enters into play. And, as we know, the moving charge produces also a magnetic field. But we have no e.m. waves.
