Is this an electromagnetic wave without the magnetic part?
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.
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.