It turns out the the previous answer had the right ingredients, but in the wrong combination. Here is a cleaner proof.

**Notation:** Let $s_j(X)$ denote the $j$-th singular value of a matrix $X$ (we assume that singular values are arranged in decreasing order). Similarly, let $\lambda_j(X)$ denote the $j$-th eigenvalue of a Hermitian matrix $X$. Let $S(X)$ denote the diagonal matrix of singular values of $X$.

**Lemma 1:** If $B \in M_n(\mathbb{C})$ is a contraction, then for any $A \in M_n(\mathbb{C})$ we have
\begin{equation*}
s_j(A) \ge s_j(AB),\qquad 1\le j \le n.
\end{equation*}

*Proof:* Since $B$ is a contraction, we have
\begin{eqnarray*}
I &\ge& BB^*\\
AA^* &\ge& ABB^*A^*\\
\lambda_j(AA^*) &\ge& \lambda_j(ABB^*A^*)\\
\lambda_j^{1/2}(AA^*) &\ge& \lambda_j^{1/2}(ABB^*A^*)\\
s_j(A) &\ge& s_j(AB).
\end{eqnarray*}

**Theorem 2:** Let $A, B \in M_n(\mathbb{C})$ be contractions. Then,
\begin{equation*}
\|I - S(A)\| \le \|I-S(AB)\| \le \|I-AB\|,
\end{equation*}
for any unitarily invariant norm $\|\cdot\|$.

*Proof:* Using Lemma 1 and that $A$ is a contraction, we have $0 \le 1 - s_j(A) \le 1 - s_j(AB)$ for all $j$. Consequently, it follows that
\begin{equation*}
\|I-S(A)\| \le \|I-S(AB)\|,
\end{equation*}
for any unitarily invariant norm. Now using a corollary of Lidkskii's majorization (see e.g., [Theorem IV.3.4 in Bha97]), it follows that the following inequality
\begin{equation*}
\|I-S(AB)\| = \|S(I)-S(AB)\| \le \|I-AB\|,
\end{equation*}
holds for all unitarily invariant norms.

**Corollary 3:** If for contractions $A$ and $B$, we have $\|AB-I\| < \epsilon$, then there exists a unitary matrix $U$ such that $\|A-U\| < \epsilon$.

*Proof:* Let $A=UP$ be the polar decomposition of $A$. Then for any unitarily invariant norm $\|\cdot\|$,
\begin{equation*}
\|A-U\|=\|P-I\|=\|S(A)-I\|.
\end{equation*}
Combining this equality with Theorem 2, the result is immediate.