Must nonunit in group algebra of free group generate proper two-sided ideal?

Question

Let $F$ be a free group and $k$ be a field. If $x$ is an element of the group algebra $k[F]$ that is not a unit (equivalently, that is not a nonzero scalar multiple of an element of $F$), must the 2-sided ideal $k[F]\,x\,k[F]$ generated by $x$ be proper? In other words, is it true that for all $y_1,\dots, y_n, z_1,\dots, z_n \in k[F]$, we have $$y_1 x z_1 + y_2 x z_2 +\dots + y_n x z_n \neq 1 ?$$

This question was asked by George Bergman. For the application he has in mind, it would actually be enough to show that if $a$ is one of the free generators of $F$, and $x$ is an element of $k[F]$ whose support consists of elements of $F$ having total degree $0$ in $a$, then $1\notin k[a]\,x\,k[F]$. (For example, $x$ might be $b + abaca^{-2}$, where $b$ and $c$ are two other members of a free generating set for $F$.) Equivalently, he would like to know that the right ideal of $k[F]$ generated by the conjugates of $x$ by the non-negative powers of $a$ is a proper ideal.

Answer 1

For one-sided ideals it's true (see for example Lemma 5.9 in this paper). The following is an unsuccessful attempt to handle two-sided ideals in the same method:

A free group is bi-orderable, so let $<$ be a bi-order on $F$.
(It means $w_1 < w_2\implies w_1 u < w_2 u, \,\,\,u w_1 < u w_2$ for every $w_1, w_2, u\in F$. See this post for details).

Assume now we have a non-unit: $x = \sum_{j=1}^d \lambda_j x_j$ for $\lambda_j\in k-\{0\}, \,\,x_1 < \ldots < x_d\in F$ with $d\ge 2$, and some $c_i\in k,\,\,\, y_i, z_i\in F$ such that $$1 = \sum_{i=1}^n c_i y_i x z_i = \sum_{i=1}^n \sum_{j=1}^d c_i \lambda_j \cdot y_i x_j z_i. $$ Let $i_{\max}, i_{\min}$ be such that $$y_{i_{\min}} x_1 z_{i_{\min}} = \min\left(\left\{ y_i x_1 z_i \right\}_{i=1}^n\right), \,\,\, y_{i_{\max}} x_d z_{i_{\max}} = \max\left(\left\{ y_i x_d z_i \right\}_{i=1}^n\right).$$ Then for every $j\neq 1$ and every $i$ we have $$y_{i_{\min}} x_1 z_{i_{\min}} \le y_{i} x_1 z_{i} < y_{i} x_j z_{i}$$ so the total coefficient of $y_{i_{\min}} x_1 z_{i_{\min}}$ is non-zero, and similarly for $y_{i_{\max}} x_d z_{i_{\max}}$.
Thus the sum cannot be only $1$ - a contradiction.

I still believe this kind of argument should work, but as is it doesn't suffice (see YCor's comment).