Here is a proof that holds under a slight addition hypothesis, namely, suppose in addition we also have $\|A^*B^*-I\| \le \epsilon$. The conclusion is also stronger than the one requested in question, as under this additional hypothesis, the claim holds for all unitarily invariant norms, not just Schatten-$p$ norms.
Let $A=UP$ be the polar decomposition of $A$. Then, we know that for unitarily invariant norms \begin{equation*} \|A-U\| \le \|A-V\| \le \|A+U\|, \end{equation*} for any unitary matrix $V$. It follows therefore that \begin{equation*} \left\|\begin{pmatrix} A & 0\\ 0 & -A^* \end{pmatrix} - \begin{pmatrix} U & 0\\ 0 & -U^* \end{pmatrix}\right\| \le \left\|\begin{pmatrix} A & 0\\ 0 & -A^* \end{pmatrix} - V^*\right\| \end{equation*} for any unitary matrix $V$. Since $B$ is a contraction, we can dilate it to a unitary matrix, e.g., \begin{equation*} V := \begin{pmatrix} B & *\\ * & -B^* \end{pmatrix}, \end{equation*} where the $*$ denotes unspecified entries. Since the norms involved are unitarily invariant, we have \begin{equation*} \|(A-U)\oplus(U-A)^*\| \le \left\|\begin{pmatrix} A & 0\\ 0 & -A^* \end{pmatrix}V - I\right\| = \|(AB-I)\oplus (A^*B^*-I)\|. \end{equation*} But notice that $s_j(A-U)=s_j(U^*-A^*)$. And we assumed that $\|AB-I\|<\epsilon$ and $|A^*B^*-I\| < \epsilon$. It thus follows that $\|A-U\| < \epsilon$.
Note: Perhaps the proof can be simplified and made more elegant while getting rid of the extra assumption.