Here is a proof that holds under a slight additional hypothesis, namely, suppose that we also have $\|A^*B^*-I\| < \epsilon$. The conclusion is also stronger than the one requested, as under this additional hypothesis, the claim holds for all unitarily invariant norms, not just Schatten-$p$ norms.

>**Theorem ([Thm. IX.7.2, [Bha97][1]])**. Let $A=UP$ be the polar decomposition of $A$. Then, for every unitarily invariant norm $\|\cdot\|$ and every unitary matrix $V$
\begin{equation*}
  \|A-U\| \le \|A-V\|.
\end{equation*}

Applying this theorem to the direct sum $A\oplus (-A^*)$ we have
\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 the unitary matrix][2]
\begin{equation*}
 V := \begin{pmatrix}
  B & (I-BB^*)^{1/2}\\
  (I-B^*B)^{1/2} & -B^*
  \end{pmatrix}.
\end{equation*}

Since the norms involved are unitarily invariant, we therefore obtain
\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 $s_j(A-U)=s_j(U^*-A^*)$, for each singular value $s_j$ for $1\le j \le n$. 
Therefore, we can conclude the following weak-majorization between the singular values:
\begin{equation*}
 s(A-U) \prec_w \max\{s_j(AB-I),s_j(A^*B^*-I)\}_{j=1}^n.
\end{equation*}
From this under our hypotheses $\|AB-I\|<\epsilon$ and $\|A^*B^*-I\| < \epsilon$, it follows that $\|A-U\| < \epsilon$.

**Note:** It should be possible to get rid of the extra assumption, but am not sure as of now.


  [1]: https://books.google.com/books?id=lh4BCAAAQBAJ
  [2]: http://en.wikipedia.org/wiki/Dilation_(operator_theory)