 Solution

The unilateral shift operator $R$ is not onto, since its outputs have first coordinate zero. The left shift operator $L$ is not injective, since $L{e}_1=0$. Both have operator norm one, and $LR=I$. ConsequentlyIf $\Vert A\Vert <1$, then $\Vert LA\Vert <1$, so the Neumann series makes $I+LA$ invertible. If $R+A$ were invertible too, the identity above would givea product of invertible operators, contradicting the nontrivial kernel of $L$. ThusThis is norm stability of noninvertibility for the unilateral right shift.