*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:**

Let $U \in \operatorname{O}_n$,$A \in M_n$ with *positive* eigenvalues, and let $\delta >0$. There exists a constant $C>0$ (independent of $U,A,\delta$) such that if $|A-U|_{op} \le \delta$, then $|U-I|_{op} \le C\delta$. (In fact one can choose $C=5n$).
 
**Why the lemma implies our required result?**

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

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

Q.E.D

**Attempted proof of the lemma:**

Assume by contradiction that $|U-I|_{op} > C\delta$. 

Since $U-I$ is normal $|U-I|_{op} = \max{|\lambda_i-1|}$ (where the $\lambda_i$ are the eigenvalues of $U$).  So, there exists an eigenvalue $\lambda=a+ib$ of $U$, such that $|\lambda-1|>C\delta$.

We claim that the distance of $\lambda$ from the **semi-positive $x$ axis** is greater than $\min \{1,\frac{1}{2}(C\delta)^2\}$.

(As commented by fedja, it is very intuitive that a point on the unit circle which is far from $1$ is far from the positive semi-axis).

**Let us separate into cases:**

**$(1)$: $a \ge 0.$**

$$ |\lambda-1|^2=2-2a>(C\delta)^2 \Rightarrow 1-a > \frac{1}{2}(C\delta)^2$$

$$b^2=1-a^2=(1-a)(1+a) \ge (1-a)^2$$ (here we used $a \ge 0$)

so $$|b| \ge 1-a > \frac{1}{2}(C\delta)^2$$

Since $a \ge 0$, it is clear that $d(\lambda,x_{\ge 0})=|b|>\frac{1}{2}(C\delta)^2$

**$(2)$: $a < 0.$** In that case $d(\lambda,x_{\ge 0})=1$