There is another way of looking at the problem that was pointed out in the 1941 paper The Electric Field Associated with a Steady Current in Long Cylindrical Conductor by <PERSON>. Like the <PERSON> paper it puts the conductor along the axis of a cylinder far away to have a return path for the current.
Since it is a steady state problem he treats it as "an interesting example of <PERSON>'s equation" Using separation of variables in cylindrical coordinate and he assumes a linear solution in z. Then he looks at the boundary conditions. 789
Following the paper (in Gaussian units).
$$V=(Az+B)(\alpha \log{r}+\beta)$$
Inside the wire $\alpha $ can be set to 0 and $\beta $ to 1.
Taking the derivative to find the electric field we see that the E-field is a constant and points along the wire and inside the wire there is no radial component to E.
$$ \frac{-dV}{dZ}= E_1 =-A_1$$
Where the subscript 1 and 2 are inside and outside the wire and we could also say if the resistance of the wire per unit length is constant that $-A_1=RI$ or $J=\sigma E$ as we expect.
Outside the wire, $$ V_2=A_2z(\alpha\log{r}+\beta)$$
Letting $\alpha$ =1 and evaluating the boundary conditions $\beta=-log(r_1)$ giving $A_2=-log(\frac{r_0}{r_1})$
So inside the wire $$V_1=-E_1z$$
outside the wire
$$V_2=-\frac{E_1z}{log\frac{r_0}{r_1}}log\frac{r}{r_1}$$
Since the potentials and the electric fields are found we can also find the surface charge using the gaussian pillbox.
$$(\frac{dV_1}{dr}){r=r_0}-(\frac{dV_2}{dr}){r=r_0}=4\pi\sigma$$
which gives
$$\sigma= \frac{E_1z}{4\pi log(\frac{r_0}{r_1})}$$
Looking at the problem this way, it agrees with <PERSON> paper statement that the surface charge is necessary to establish the potentials outside the wire.
Also in steady state, if you can find the potentials inside and outside the wire you can find the surface charge density. There are some pedagogical papers in the American Journal of physics that do this numerically. In doing this you can also see how the charge is distributed at bends in the wire. If appropriate time steps are taken one can see the transient of how the charge moves around and relaxes when a switch is thrown. 903