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.