If the positive and negative charges are moving at different velocities, then their contributions to the current density will be different, even if their charge densities are equal and opposite.
For example, in a normal wire, we can usually model the electrons as moving and the positive charges as stationary (though this is material-dependent and in particular doesn't hold for semiconductors). The charge density of positive charges is $\rho$ and the charge density of electrons is $-\rho$, so the total charge density is $\rho-\rho=0$. But the electrons are moving at velocity $-\vec{v}$ (where $\vec{v}$ points in the direction of the electric field) and the protons are effectively stationary, so the total current density is $\rho\vec{0}-\rho(-\vec{v})=\rho\vec{v}$, which is nonzero. You can add the current densities like this because Maxwell's equations are linear, and therefore obey the principle of superposition.
For another example, suppose we somehow had a wire made out of electrons and positrons of equal charge density (neglecting annihilation or any of the more interesting physics that this would imply). When an electric field is applied to this wire, the positrons will move in the direction of the electric field and the electrons will move in the opposite direction. Since the electron and positron charge have the same absolute value, the magnitude of the force on each species will be the same, and they will move at the same speed. Once again, the charge density is zero, since the positrons have charge density $\rho$ and the electrons have charge density $-\rho$. For the current density, we define $\vec{v}$ to be pointing in the same direction as the electric field; then the positrons have charge density $\rho$ and are moving with velocity $\vec{v}$, and the electrons have charge density $-\rho$ and are moving with velocity $-\vec{v}$, so the total current density is $\rho\vec{v}-\rho(-\vec{v})=2\rho\vec{v}$. So even when the charge carriers move at the same speed, the current density can still be nonzero if their velocity vectors are pointing in different directions.
This form of Maxwell's Equations should cover metals dielectrics and most other media. Focus on the following
$$
\begin{align}
\boldsymbol{\nabla}.\mathbf{D}&=\rho_f \\
\boldsymbol{\nabla}\times\mathbf{H}&=\mathbf{J}_f+\partial_t\mathbf{D}
\end{align}
$$
Now take divergence of the second equation, and bear in mind that divergence of a curl is zero:
$$
\boldsymbol{\nabla}.\boldsymbol{\nabla}\times\mathbf{H}=\boldsymbol{\nabla}.\mathbf{J}_f+\boldsymbol{\nabla}.\partial_t \mathbf{D}=0
$$
You can then swap and substitute $\boldsymbol{\nabla}.\partial_t \mathbf{D}=\partial_t \boldsymbol{\nabla}.\mathbf{D}=\partial_t \rho_f$
So the continuity equation is a direct consequence of Maxwell's equations. Verifying it is akin to verifying Maxwell's equations. No need to mess around with thing wires and loops, the above prescription applies to most media and even mixed domains, i.e. with different media.
The above analysis tells you that you cannot talk about charge density and current density separately.
I am not sure what proof you are expecting for infinitely thin wire. There I would simply state current density as:
$$
\mathbf{J}\left(\mathbf{r}\right)=\alpha\int ds'\: \boldsymbol{\mathcal{\dot{R}}}\left(s'\right)\delta^{\left(3\right)}\left(\mathbf{r}-\boldsymbol{\mathcal{R}}\left(s'\right)\right)
$$
Where $\boldsymbol{\mathcal{R}}\left(s\right)$ is the trajectory of your wire, parametrized by arc-length $s$. Assuming $\boldsymbol{\mathcal{\dot{R}}}.\boldsymbol{\mathcal{\dot{R}}}=\mathcal{\dot{R}}^2=const$
Value of $\alpha$ follows from:
$$
\int_S{d^2}r\: \mathbf{\hat{n}}.\mathbf{J}=I\Delta\left(s\in S\right)\cos\theta_\mathcal{\dot{R}}=\alpha\mathcal{\dot{R}}^2\,\Delta\left(s\in S\right)
$$
Where the integral above is over the surface area $S$ of the dot-product of the current density with the surface normal $\mathcal{\hat{n}}$. Quantity $\Delta\left(s\in S\right)=1$ is the wire only goes through the surface area $S$ once. $\cos\theta_{\mathcal{\dot{R}}}$ is the cosine of the angle between the wire and the surface normal. $I$ is the current in the wire. Equation above only makes sense if wire goes through $S$ once or not at all.
From this you can then extract the charge density by taking divergence. I think, you will find that once you take divergence the derivative (of the divergence) and the integral ($\int ds'$) will cancel out, and you will get zero charge density for closed-loop wires
ADDENDUM
Lets see how certain integrals/derivatives will transform
Let us restrict our attention to a region, $\left(x,y,z\right)\in\Omega$ and $s\in\left(s_0,s_1\right)$, where $z$ has a one-to-one relationship with arc-length $s$. More specifically, $\mathcal{R}_z\left(s\right)$ gives the z-coordinate of the arc at all points. Within the region we are considering, let $\mathcal{R}_z^{-1}\left(z\right)$ be defined.
Irrespective of where $z$ is the component of the arc-tangent (of the wire) in the direction of $z$ is:
$$
\frac{d\mathcal{R}_z}{ds}=\mathbf{\hat{z}}.\frac{d\boldsymbol{\mathcal{R}}}{ds}=\mathbf{\hat{z}}.\boldsymbol{\dot{\mathcal{R}}}
$$
Next note that within $\Omega$ for any function $f=f\left(x,y,z\right)$:
$$
\begin{align}
\int^b_a dz\, f\left(x,y,z\right)&=\int^{\mathcal{R}_z^{-1}\left(b\right)}_{\mathcal{R}_z^{-1}\left(a\right)} \frac{dz}{ds} ds f\left(x,y,\mathcal{R}_z\left(s\right)\right)=\int^{\mathcal{R}_z^{-1}\left(b\right)}_{\mathcal{R}_z^{-1}\left(a\right)} \frac{d\mathcal{R}_z}{ds} ds f\left(x,y,\mathcal{R}_z\left(s\right)\right)\\
&=\mathbf{\hat{z}}.\int^{\mathcal{R}_z^{-1}\left(b\right)}_{\mathcal{R}_z^{-1}\left(a\right)} \boldsymbol{\mathcal{\dot{R}}} f\left(x,y,\mathcal{R}_z\left(s\right)\right)ds
\end{align}
$$
Where I simply replaced the integration variable $z\to s$ where $z=\mathcal{R}_z\left(s\right)$. The same trick will work in volume integrals within $\Omega$. The the transform would be:
$$
\begin{align}
x &\to x \\
y&\to y \\
z &\to s \\
\end{align}
$$
With the Jacobian $\left|\frac{\partial\left(x,y,z\right)}{\partial\left(x,y,s\right)}\right|=\mathbf{\hat{z}}.\boldsymbol{\mathcal{\dot{R}}}$
It then follows that (within $\Omega$):
$$
\begin{align}
\mathbf{\hat{z}}.\mathbf{J}\left(\mathbf{r}\right)&=\alpha\mathbf{\hat{z}}.\int ds'\: \boldsymbol{\mathcal{\dot{R}}}\left(s'\right)\delta^{\left(3\right)}\left(\mathbf{r}-\boldsymbol{\mathcal{R}}\left(s'\right)\right)=\alpha\int dz'\: \delta^{\left(3\right)}\left(\left(\begin{array}\\x\\y\\z\end{array}\right)-\left(\begin{array}\\\mathcal{R}_x\left(\mathcal{R}_z^{-1}\left(z'\right)\right)\\\mathcal{R}_y\left(\mathcal{R}_z^{-1}\left(z'\right)\right)\\z'\end{array}\right)\right)=\\
&=\alpha \:\delta\left(x-\mathcal{R}_x\left(\mathcal{R}_z^{-1}\left(z\right)\right)\right)\:\delta\left(y-\mathcal{R}_y\left(\mathcal{R}_z^{-1}\left(z\right)\right)\right)
\end{align}
$$
From here it should be relatively easy to derive:
$$
\int_{x_0}^{x_1} dx \int_{y_0}^{y_1} dy \int_{-l/2}^{l/2} dz\: \mathbf{\hat{z}}.\mathbf{J}\left(\mathbf{r}\right)=l\,\alpha
$$
Assuming that $\left(x_0,\,x_1\right)\times\left(y_0,y_1\right)\times\left(-l/2,l/2\right)$ contain the wire and are within $\Omega$. I think this is what you were after.
Another interesting thing to try is:
$$
\begin{align}
\boldsymbol{\nabla}.\mathbf{J}&=\int^{s_b}_{s_a} ds'\: \boldsymbol{\mathcal{\dot{R}}}\left(s'\right).\boldsymbol{\nabla}\delta^{\left(3\right)}\left(\mathbf{r}-\boldsymbol{\mathcal{R}}\left(s'\right)\right)=\int^{s_b}_{s_a} ds'\: \frac{d\boldsymbol{\mathcal{R}}}{ds'}.\boldsymbol{\nabla}\delta^{\left(3\right)}\left(\mathbf{r}-\boldsymbol{\mathcal{R}}\left(s'\right)\right)
\\
&=-\int^{s_b}_{s_a} ds'\: \frac{d}{ds'}.\delta^{\left(3\right)}\left(\mathbf{r}-\boldsymbol{\mathcal{R}}\left(s'\right)\right)=\delta^{\left(3\right)}\left(\mathbf{r}-\boldsymbol{\mathcal{R}}\left(s_a\right)\right)-\delta^{\left(3\right)}\left(\mathbf{r}-\boldsymbol{\mathcal{R}}\left(s_b\right)\right)
\end{align}
$$
Clearly, if $s_a=s_b$ as would be in the case of a closed loop, the divergence of the current density would vanish.
Best Answer
Net charge density is independent of current density: