Suppose $a(L)$ and $b(L)$ are series of negative powers of the lag operator, that is $$a(L) = \sum_{j=1}^\infty a_jL^{-j},\quad\text{and}\quad b(L) = \sum_{j=1}^\infty b_jL^{-j}.$$ Also suppose that $||a(L)||<1$ with the operator norm, and suppose that $c$ is a scalar with $|c|<1$.
Which assumption must $b(L)$ satisfy to guarantee that $$\left\Vert \frac{b(L)+c}{1-a(L)} \right\Vert <1 ?$$
Obs: If $$\left\Vert \frac{b(L)+c}{1-a(L)} \right\Vert \leq \frac{|| b(L)+c||}{||1-a(L)||},$$ then it is easy to see that $$ ||a(L)||+||b(L)||+c<1$$ would be sufficient.
Obs2: By 'lag operator' I mean the shift operator with a unit shift, so that $||L||=1$.
Since the lag operator is commutative, we can safely read \begin{align*} \frac{b(L)+c}{1-a(L)} = [b(L)+c][1-a(L)]^{-1} \end{align*} Taking norms, \begin{align*} \|[b(L)+c][1-a(L)]^{-1}\| \le \|b(L)+c\|\cdot\|[1-a(L)]^{-1}\| \end{align*} Then, as @mzp pointed out from this post with $T = 1 - a(L)$, we have \begin{align*} \|[1-a(L)]^{-1}\| \le \frac{1}{1 -\|a(L)\|} \end{align*} And so \begin{align*} \|[b(L)+c][1-a(L)]^{-1}\| \le \frac{\|b(L)\|+c}{1 - \|a(L)\|} \end{align*} A sufficient condition is therefore $\frac{\|b(L)\|+c}{1 - \|a(L)\|} < 1$ or $\|b(L)\| < 1 - \|a(L)\| - c$, which is exactly the condition @mzp presents under Obs!