Is this an electromagnetic wave without the magnetic part? Jefimenko's Equations are:
$$
\begin{align}
&\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{\rho(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t}\right)(\mathbf{r}-\mathbf{r}') - \frac{1}{|\mathbf{r}-\mathbf{r}'| c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \mathrm{d}^3 \mathbf{r} \\
&\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \times (\mathbf{r}-\mathbf{r}') \,\mathrm{d}^3 \mathbf{r} \\
& \mbox{where the retarded time is:  }t_r = t - \frac{|\mathbf{r}-\mathbf{r}'|}{c}
\end{align}
$$

I noticed the following: In the E-field, the second and third terms fall at $1/r$. Meaning, they are radiative terms. That is, electromagnetic radiation. Analogously, In the B-field, the second term falls at $1/r$, thus radiative. However, in the B-Field, there is no radiative term depending on $\partial\rho / \partial t$.
With that in mind, consider the following situation:
$$
\frac{\partial\mathbf J}{\partial t} = 0
\quad\quad\mbox{and}\quad\quad
\frac{\partial\rho}{\partial t} \neq 0
$$
Then, the magnetic field will not be time varying, and won't be radiative. In this situation, the only radiative fields would be electric. Meaning, an electromagnetic radiation without the magnetic part! How is this possible? Is there a mistake somewhere? What prevents it from happening?

An approach was to allow time-varying $\rho$ with charge conservation by means of constant not-null-everywhere $\mathbf J(\mathbf r)$. By continuity equation, one can have a time varying $\rho$ one wants, by simply choosing $\mathbf J$ cleverly. I haven't made further progress with this.
$$
\nabla\cdot\mathbf J + \frac{\partial\rho}{\partial t} = 0
\quad\implies\quad
\frac{\partial\rho}{\partial t} = -\nabla\cdot\mathbf J
$$
 A: Arguably, it might be, but the sources you've provided cannot be sustained for long: since the current doesn't change in time, you have
$$
\frac{\partial^2\rho}{\partial t^2} = -\nabla \cdot\frac{\partial \mathbf J}{\partial t} = 0,
$$
i.e. if the charge density is increasing with time, then it is doing so linearly with time and without any way to stop, so you will necessarily end up with regions of arbitrarily high charges over sufficiently long times, and that is going to take an unbounded amount of energy.
If you're OK with that, then yeah, you can go on and find out whatever weird properties the emitted field has, but it's not really something that's broadly considered to be a physically allowed situation.
A: To add to Emilio Pisanty's answer: for the situation you are considering, Jefimenko's equations simplify to
$$
\begin{align}
\mathbf{E}(\mathbf{r}, t) &= \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{\rho(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r} \\
\mathbf{B}(\mathbf{r}, t) &= \frac{\mu_0}{4 \pi} \int \frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} \times (\mathbf{r}-\mathbf{r}') \,\mathrm{d}^3 \mathbf{r} \\
\end{align}
$$
As Emilio points out, if ${\bf J}$ does not depend on time then the continuity equation requires that $\rho$ depends linearly on time. Writing it as $\rho({\bf r'}, t) = a({\bf r'}) + b({\bf r'})\, t$ gives
$$
\begin{align*}
\mathbf{E}(\mathbf{r}, t) &= \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{a({\bf r'}) + b({\bf r'})\, t_r}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{b({\bf r'})}{|\mathbf{r}-\mathbf{r}'|^2 c}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r} \\
&= \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{a({\bf r'}) + b({\bf r'})\, \left(t - \frac{|{\bf r} - {\bf r'}|}{c} \right)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{b({\bf r'})}{|\mathbf{r}-\mathbf{r}'|^2 c}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r} \\
&= \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{a({\bf r'}) + b({\bf r'})\,  t}{|\mathbf{r}-\mathbf{r}'|^3}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r} \\
&= \frac{1}{4 \pi \epsilon_0} \int \left[ \frac{\rho({\bf r', t})}{|\mathbf{r}-\mathbf{r}'|^3} (\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r}.
\end{align*}
$$
All retardation effects cancel out, and the electromagnetic fields are given by the usual Coulomb and Biot-Savart laws applied to the instantaneous rather than retarded sources!  In particular, they fall of like $1/r^2$, and are not radiative.
