You are roughly correct. However, you must be careful because the surface you would choose for finding the magnetic field from $\mathbf{J} \cdot \mathrm{d}\mathbf{S}$ is NOT the same surface you would use to find the electric field.
The concept you want to solve this problem is self-inductance. Defining the magnetic flux $\Phi=\int\mathbf{B} \cdot \mathrm{d}\mathbf{S}$ and the electromotive force $\varepsilon = \oint\mathbf{E} \cdot \mathrm{d}\mathbf{S}$, we can rewrite the Maxwell-Faraday equation as
$$\varepsilon = -\frac{\mathrm{d}\Phi}{\mathrm{d}t}$$
and note that the current produced is
$$I=\frac{V+\varepsilon-\frac{Q}{C}}{R}$$
i.e., the sum of voltages around the whole circuit.
In general the total magnetic flux $\Phi$ through a circuit will depend in some complicated way on the geometry of the circuit, and it is hard to solve except in a few simple cases like solenoids. However, we can see from the second Maxwell equation that it will always be proportional to the current. (The second term is zero since there is no electric field perpendicular to the circuit in this problem.) Let's call the complicated geometric dependence the self-inductance $L$, and rewrite the second Maxwell equation as
$$ \Phi = LI$$
You can now write that
$$ RI = V-L\frac{\mathrm{d}I}{\mathrm{d}t}-\frac{Q}{C}$$
Observing that $I=\frac{\mathrm{d}Q}{\mathrm{d}t}$ lets you rewrite the expression in terms of a single differential equation with one variable $Q$. Once you have the solutions for $Q(t)$, you can find the behavior of, say, $\varepsilon$ through the equations we already defined and appropriate initial conditions. If you have not seen equations of this type before, it may help to search for "damped harmonic oscillator."
Your proposed vector potential diverges at $r = 0$. This may not seem like an insurmountable problem—after all, we see infinite potentials all the time for things like point charges & line currents, right? It turns out, in fact, that there's an infinite flux hiding in this problem.
Consider a a loop of radius $\epsilon$ about the origin in the $xy$-plane. The magnetic flux through this surface is equal to the line integral of $\vec{A}$ around this curve, which is:
$$
\Phi = \oint \vec{A} \cdot d\vec{l} = \int_0^{2 \pi} \frac{B_z}{2} \left( \epsilon - \frac{R^2}{\epsilon} \right) \hat{\phi} \cdot (\epsilon \,d\theta \hat{\phi}) = \pi B_z \left(\epsilon^2 - R^2 \right)
$$
The first term is what we would expect from the infinite solenoid; but what's that second term doing there? It's independent of $\epsilon$, which means that the loop is "catching" this flux no matter how small the loop is. In other words, your proposed vector potential hides an infinitely strong, dense magnetic flux concentrated on the $z$-axis, of magnitude $-\pi B_z R^2$. You can think of this as the limit of two nested solenoids, with opposing currents and radii $R_i < R_o$, in the limit as $R_i \to 0$. In terms of delta-functions, the curl of this vector field would be
$$
\vec{B} = - \pi B_z R^2 \delta(x) \delta(y) \hat{z} + \begin{cases} B_z \hat{z} & r < R \\ 0 &r > R \end{cases}.
$$
The reason this doesn't show up when you take the curl using the formulas in the endpapers of Griffiths or Jackson, by the way, is that those formulas are only guaranteed to work for points where the coordinates are not singular. Roughly speaking, a singular point of a coordinate system is any point where one or more of the basis vectors are not well-defined. At $r = 0$ in a cylindrical coordinate system, the basis vectors $\hat{r}$ and $\hat{\phi}$ are not well-defined, and so the coordinates are singular there and the usual vector calculus formulas require caution to use.
Best Answer
The magnetic vector potential isn't the induced electric field. However, in the case that the net charge density $\rho = 0$
The induced electric field can be expressed purely in terms of the magnetic vector potential.
$\vec{E} = - \frac{\partial \vec{A}}{\partial t}$
Infact, given the net charge density is not zero, this equation can also be used to find the induced solenoidal component of the E field, provided we are in the coulomb gauge.