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

We shall need the following preliminary results:

**Lemma 1:**
Let $\lambda \in \mathbb{S}^1$. Then, the distance of $\lambda$ from $1$ is not greater than twice its distance to the set of non-negative reals.

**Proof of lemma 1:**

Denote $\lambda = a+ib$. Note that $|\lambda-1|= \sqrt{2-2a}$.

**We separate into two cases:**

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

Since $a \ge 0$, it is clear that $d(\lambda,x_{\ge 0})=|b|$, so

$$2d(\lambda,x_{\ge 0}) \ge |\lambda-1| \iff 2|b| \ge \sqrt{2-2a} \iff 4b^2 \ge 2-2a \iff $$

$$ 2-2a^2=2b^2 \ge 1- a \iff a+1-2a^2 \ge 0$$

This holds since the L.H.S equals $a(1-a)+(1-a^2)$ which is a sum of two non-negative numbers. (Remember $0\le a \le 1$).



**$(2)$: $a < 0.$** In that case $d(\lambda,x_{\ge 0})=1$, so the inequality becomes $2 \ge |\lambda-1|$ which is trivial (The diameter of the unit circle is $2$).
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**Lemma 2:** 

Let $A_t, \, \,(t\in[0,1])$ be a continuous family of matrices and $K$ a compact set on the complex plane (with a continuous **connected** boundary). If the boundary of $K$ contains no eigenvalues of $A_t$ for all $t\in(0,1)$, then all $A_t$ have the same number of eigenvalues in $K$. 

**Proof of lemma 2:**

Let $P_t$ be the characteristic polynomial of $A_t$. Then by the assumption on the eigenvalues of $A_t$, $P_t|_{\partial K} \neq 0$. 

Since $A_t$ depends continuously on $t$, and the characteristic polynomial of a matrix depends continuously on its entries, $P_t$ depends continuously on $t$. 

Fix $t \in (0,1]$. We want to show $P_t,P_0$ have the same number of roots in $K$.  By the [Rouch Theorem][4], this holds if $$ |P_t(z)-P_0(z)| < |P_t(z)|+|P_0(z)| $$ for every $z \in \partial K$.

Assume by contradiction that for some $z \in \partial K$,$|P_t(z)-P_0(z)| = |P_t(z)|+|P_0(z)| $. 

Then $P_t(z)\cdot \overline{P_0(z)} \le 0$. Look at the function $s \to P_s(z)\cdot \overline{P_0(z)}$ defined on $[0,t]$; It is positive at $s=0$, and non-positive at $s=t$. By continuity, there is some $s$ such that $P_s(z)\cdot \overline{P_0(z)}=0$, so $P_s(z)=0$ which is a contradiction.
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**Back to the main proposition:**

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 3:**

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 lemma 3 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 3:**

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 $\lambda$ of $U$, such that $|\lambda-1|>C\delta$.

Since $\lambda \in \mathbb{S}^1$ lemma (1) implies that the distance of $\lambda$ from the semi-positive $x$ axis is greater than $\frac{1}{2}C\delta$.

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[4]:https://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem#Symmetric_version