*I am trying to fill in some details in fedja's answer:*

We want to prove $$(1) \, \, |AB-O_{AB}| \ge c|AB-O_AO_B|$$ for some $1>c>0$.
Let $A=O_AP_A,B=O_BP_B$ be the polar decompositions of $A,B$. Then $$AB=O_AO_B(O_B^TP_AO_B)P_B=O_AO_BXY,$$ where we Denote $$X=O_B^TP_AO_B,Y=P_B \, \text{  (both are symmetric positive definite) }$$

Then $(1)$ becomes:$$ |O_AO_BXY-O_{AB}| \ge c|O_AO_BXY-O_AO_B|=c|XY-I|$$

(The last equality holds whether we use the Frobenius norm, or the operator norm, since both are invariant under multiplication by orthogonal matrices).

Denoting $U_{A,B}=(O_AO_B)^{-1}O_{AB}$, and using again the orthogonal invariance of the norm we get that $(1)$ is equivalent to $$ |XY-U_{A,B}| \ge c|XY-I|$$

Note that $XY$ similar to $X^{1/2}YX^{1/2}$, so all its eigenvalues are real positive.

Thus, it suffices to prove the following lemma:

**Lemma:**

Let $U \in \operatorname{O}_n$,$A \in M_n$ with *positive* eigenvalues. There exists a constant $C>0$ (independent of $U,A$) such that if $|A-U| \le \delta$, then $|U-I| \le C\delta$. (In fact one can choose $C=5n$).
 
(Here $|\cdot|$ is the **operator norm**?)

**Why the lemma implies our required result?**

Taking $\delta=|A-U|$ we get: $$|A-I| \le |A-U|+|U-I| \le (C+1)|A-U|$$.
Putting $A=XY,U=U_{A,B}$ this becomes:

$$ |XY-I| \le (C+1)|XY-U_{A,B}|$$

Q.E.D

**Attempted proof of the lemma:**

Assume by contradiction that $|U-I| > C\delta$. We claim this implies that there exist an eigenvalue of $U$ that is at distance at least $\frac 12 C\delta$ from the positive semi-axis. Explicitly, there exists $\lambda=a+ib$ such that $|b| > \frac 12 C\delta$.

*I do not understand why there is such an eigenvalue $\lambda$*. Moreover, we can't take in our application $\delta$ to be arbitrarily small, so it can happen that $\frac 12 C\delta > 1$ which contradicts the fact $|\lambda|=1$ (since it is an eigenvalue of an orthogonal matrix).