Suppose that $(A_n^{-1})$ is bounded. Using the identity $a^{-1}-b^{-1}=a^{-1}(b-a)b^{-1}$ and the fact that $(A_n)$ is a Cauchy sequence, it follows that $(A_n^{-1})$ is a Cauchy sequence. Since $L(B)$ is complete, there exists an operator $T$ such that $A_n^{-1}\to T$. Taking the limit of $A_nA_n^{-1}=A_n^{-1}A_n = I$ shows that $T=A^{-1}$.
Rearranging the same identity, $(I+a^{-1}(b-a))b^{-1}=a^{-1}$. If $A$ is invertible, then $(I+A^{-1}(A_n-A))A_n^{-1}=A^{-1}$. Since $T_n:=A^{-1}(A_n-A)\to 0$, $I+T_n$ is eventually invertible, with $(I+T_n)^{-1}=\sum\limits_{k=0}^{\infty}(-T_n)^k$, and $\|(1+T_n)^{-1}\|\leq \dfrac{1}{1-\|T_n\|}\to 1$. Thus, for $n$ sufficiently large, $A_n^{-1}=(I+T_n)^{-1}A^{-1}$, and this implies that $(A_n^{-1})$ is bounded.