= Solution
For <Fourier stability analysis>, take spatial modes $e^{im\theta}$ and put
$$
D(\theta)=ae^{-i\theta}+b+ce^{i\theta},\qquad
N(\theta)=d+ee^{i\theta},\qquad G(\theta)=N(\theta)/D(\theta).
$$
A bounded implicit step requires $D$ to stay away from zero, and <von Neumann stability analysis> requires $|G|\leq1$ for every frequency. Expanding the squared moduli and using $\cos2\theta=2\cos^2\theta-1$ gives
$$
\begin{aligned}
|D|^2&=a^2+b^2+c^2+2b(a+c)\cos\theta+2ac\cos2\theta,\\
|N|^2&=d^2+e^2+2de\cos\theta,\\
\boxed{|D|^2-|N|^2}&=\boxed{\mu(\mu-2)(\mu-1)(\mu+1)(1-\cos\theta)^2.}
\end{aligned}
$$
Consequently, apart from denominator invertibility, the necessary and sufficient sign condition is
$$
\mu(\mu-2)(\mu-1)(\mu+1)\geq0.
$$
Its real solution set is $(-\infty,-1]\cup[0,1]\cup[2,\infty)$.
To check the denominator rather than assume it, note
$$
\operatorname{Im}D=(1-2\mu)\sin\theta,\qquad
D(0)=3,\qquad D(\pi)=1+2\mu-2\mu^2.
$$
If $\mu\ne1/2$, a zero can occur only at zero or the Nyquist frequency. The zero-frequency value is nonzero, and the Nyquist zeros occur only at $\mu=(1\pm\sqrt3)/2$, both outside the proposed stable intervals. At $\mu=1/2$, $D=9/4+(3/4)\cos\theta\geq3/2$, also nonzero. Continuity on the frequency circle now gives a positive lower bound on $|D|$ for each allowed $\mu$.
The implicit convolution operator is therefore boundedly invertible on the square-summable grid sequences. By the <Plancherel theorem>, $|G|\leq1$ gives
$$
\boxed{\|u^{n+1}\|_{\ell^2_h}\leq\|u^n\|_{\ell^2_h},\qquad
\mu\in(-\infty,-1]\cup[0,1]\cup[2,\infty).}
$$
Outside these intervals the modulus difference is negative at every nonzero frequency where $D$ is nonzero. There is an interval on which $|G|>1$, giving arbitrarily large growth of powers of the step operator, so the method is unstable; a zero of $D$ additionally makes the step ill posed.
For the physical choice $k,h>0$, the stable range is \b[$0<\mu\leq1$ or $\mu\geq2$]; zero is the limiting zero-step case. The exact-shift values have unit gain and are included.
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