Magnetic Field in a Leaky Capacitor

Assuming that you mean to say that flow of energy can only be radially outwards due to symmetry, and since $vec E$ already points in that direction, you can't have $vec S$ pointing in that direction, so $vec S$ must be $0$, and hence $vec B$ must be parallel to $vec E$, but this violates divergenceless-ness of $vec B$, then I think your method is correct.

Here is an alternate method:

Let the charge on the inner conductor be $Q(t)$ The current density at a distance $r$ from the centre is $$vec j = -4pi r^2frac{dQ}{dt} hat r$$

The electric field there is $vec E = frac{Q}{4pi epsilon_0 r^2} hat r$, so $$frac{dvec E}{dt} = frac{1}{4piepsilon_0 r^2}frac{dQ}{dt} hat r$$

Now, we have $nabla cdot vec{B} = 0$ and also $nablatimesvec B = mu_0 vec j + mu_0 epsilon_0 frac{dvec E}{dt} = 0$ (this is valid outside the outer surface as well since $frac{dvec{E}}{dt}$ and $vec{j}$ are $0$ there anyways). Using the boundary condition that $vec{B}$ goes to $0$ as $r$ goes to infinity, we get $vec B = 0$