= Solution
Use $z=\lambda_z>0$. In cylindrical orthonormal bases the <principal stretches> are $r_{,R}$, $r/R$ and $z$. <Incompressibility> gives
$$
r_{,R}\frac rR z=1,\qquad
\frac{d(r^2)}{dR}=\frac{2R}{z}.
$$
Integrate from the inner surface to obtain the <inflation and extension of an incompressible tube> kinematics
$$
r^2=a^2+\frac{R^2-a_0^2}{z},\qquad
\boxed{\lambda=\frac rR=z^{-1/2}
\left(1+\frac{za^2-a_0^2}{R^2}\right)^{1/2}.}
$$
In particular $b^2=a^2+(b_0^2-a_0^2)/z$. The radial <principal stretch> is $1/(z\lambda)$, so the reference-volume <strain energy density> is $\widehat W(\lambda,z)$ with the three arguments $(1/(z\lambda),\lambda,z)$.
For reference length $L_0$, the stored energy is $L_0\mathcal E$, where
$$
\mathcal E(a,z)=2\pi\int_{a_0}^{b_0}\widehat W(\lambda(R;a,z),z)R\,dR.
$$
Let $N_{\rm wall}$ denote the total axial resultant acting on the material annulus. The work of the inner <pressure> on the cylindrical wall is $2\pi azL_0p\dot a$, and axial work is $N_{\rm wall}L_0\dot z$. Reversible <virtual work> therefore says
$$
2\pi azp\dot a+N_{\rm wall}\dot z
=\mathcal E_{,a}\dot a+\mathcal E_{,z}\dot z.
$$
The independent derivatives of the circumferential stretch are
$$
\lambda_{,a}=\frac{a}{\lambda R^2},\qquad
\left.\lambda_{,z}\right|_a
=-\frac{R^2-a_0^2}{2\lambda z^2R^2}.
$$
Equating the coefficients of $\dot a$ and $\dot z$ proves
$$
\boxed{p=\frac1z\int_{a_0}^{b_0}
\widehat W_{,\lambda}\frac{dR}{\lambda R},}
$$
and the corresponding axial resultant is
$$
\boxed{N_{\rm wall}=2\pi\int_{a_0}^{b_0}
\left[\widehat W_{,z}R
-\widehat W_{,\lambda}\frac{R^2-a_0^2}{2\lambda z^2R}\right]dR.}
$$
Here $\widehat W_{,z}$ holds $\lambda$ fixed. If $N$ means the independently applied load on pressurized closed ends, the <pressure> also does axial end-cap work $\pi a^2pL_0\dot z$. Equivalently its complete work is $p\,d(\pi a^2zL_0)/dt$. In that convention,
$$
\boxed{N=N_{\rm wall}-\pi a^2p.}
$$
If $N$ includes the end-pressure resultant, it is $N_{\rm wall}$ itself. Stating this distinction is necessary because the ends are not specified; the <pressure> formula is the same in both conventions.
For the specified <Mooney-Rivlin solid> convention, differentiating the constrained <strain energy density> gives
$$
\widehat W_{,\lambda}
=(\mu_1-\mu_2z^2)(\lambda-z^{-2}\lambda^{-3}).
$$
Thus
$$
p=z^{-1}(\mu_1-\mu_2z^2)
\int_{a_0}^{b_0}(1-z^{-2}\lambda^{-4})\frac{dR}{R}.
$$
Put $c=za^2-a_0^2$. If $c\ne0$, $z\lambda^2-1=c/R^2$ yields $dR/R=-z\lambda\,d\lambda/(z\lambda^2-1)$. The factor cancels because
$$
\frac{\lambda(1-z^{-2}\lambda^{-4})}{z\lambda^2-1}
=\frac1{z\lambda}+\frac1{z^2\lambda^3}.
$$
The limits are $\lambda_a=a/a_0$ and $\lambda_b=b/b_0$, so direct integration gives
$$
\boxed{p=(\mu_1z^{-2}-\mu_2)
\left[z\log\frac{\lambda_a}{\lambda_b}
-\frac12(\lambda_a^{-2}-\lambda_b^{-2})\right].}
$$
For $c=0$, all circumferential stretches equal $z^{-1/2}$ and the original integrand vanishes, so this formula still holds, with $p=0$. This also verifies its continuous limiting value without dividing by zero.
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