= Solution
The relevant one of the <characteristic polynomials of a linear multistep method> is
$$
\rho(w)=w^3+(a-1)w-a=(w-1)(w^2+w+a).
$$
The coefficients on the right are chosen to give third order, so <consistency of a numerical method> is already available. By the <Dahlquist equivalence theorem>, convergence with convergent starting values is equivalent to <zero-stability>: every root of $\rho$ must satisfy $|w|\leq1$, and every root on the unit circle must be simple.
If $a<0$, the quadratic is negative at $w=-1$ and tends to $+\infty$ as $w\to-\infty$, so it has a root less than $-1$. If $a>1$, its root product is $a$, so at least one root has modulus greater than one. For $0\leq a\leq1/4$, its real roots $(-1\pm\sqrt{1-4a})/2$ lie in $[-1,0]$. For $1/4<a\leq1$, its conjugate roots have modulus $\sqrt a\leq1$.
At $a=0$, the three roots are $1,0,-1$, with the unit-modulus roots simple. At $a=1$, they are $1,e^{2\pi i/3},e^{-2\pi i/3}$, again simple. The repeated quadratic root at $a=1/4$ is $-1/2$, strictly inside the unit circle, which the <root condition for a multistep method> permits. Therefore
$$
\boxed{\text{the method is convergent exactly for }0\leq a\leq1.}
$$
This is the <root stability of a cubic multistep polynomial>; oscillatory parasitic roots at the endpoints do not invalidate convergence when their starting amplitudes tend to zero.
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