You are asking whether the following holds.

**Claim.** Let $F$ be a field. Then the free associative algebra $F\langle x,y\rangle$ is a $GE_2$-ring in the sense of P. M. Cohn [2].

The answer is **yes**. Indeed, free associative algebras over a field are more generally $GE$-rings [Theorem 3.4, 2]. This fact has been used by Bass [1] to prove that a finitely generated projective module over a free algebra with coefficients in a PID is free.
By contrast, free commutative algebras over a field, i.e., polynomial rings over a field **aren't** $GE$-rings if the number of indeterminates exceeds $1$ [Proposition 7.3, 2]. (I have made this remark in another MO post).

We recall below the definition of a $GE$-ring in the sense of P. M. Cohn and we give subsequently a proof of the claim.

Let $R$ be an associative ring with identity and let $E_n(R)$ be the subgroup of $GL_n(R)$ generated by matrices obtained from the identity matrix by replacing an off-diagonal entry by some $r∈R$. Let $D_n(R)$ be the group of invertible diagonal matrices; note that it normalizes $E_n(R)$.

A ring $R$ is termed a $GE_n$-ring if $GL_n(R)=E_n(R)D_n(R)$.
The ring $R$ is a $GE$-ring in the sense of P. M. Cohn if it is a $GE_n$-ring for every $n > 1$.

Let us turn to the proof of the claim. Let $S$ be the free monoid on $\{x, y\}$ endowed with the shortlex order, i.e., words are ordered according to word length and words of the same length are ordered lexicographically. Then $S$ is a bi-ordered monoid and it is not difficult to show that $F\langle x,y \rangle = F[S]$ is an integral domain with unit group $F^{\times}$. Let $\varphi: S \rightarrow \omega$ be an order-preserving bijection. The following lemma is instrumental.

**Lemma.** Let $R = F\langle x,y \rangle$ and let $a_1, a_2 \in R$. Assume there are $b_1, b_2 \in F\langle x,y \rangle$, not both zero, such that $a_1b_1 + a_2 b_2 = 0$. Then we can find $E \in E_2(R)$ and $a \in R$ such that $(a_1, a_2)E = (a, 0)$.

*Proof*. For $r \in R$, write $r = \sum_{w \in S} r_w w$, where each $r_w \in F$ and $\text{supp}(r) \Doteq \{w \in S \,\vert\, r_w \neq 0\}$ is a finite subset of $S$. Since $R$ is an integral domain, we can assume, without loss of generality, that none of $a_1, a_2, b_1$ and $b_2$ is zero. Setting $n(r) \Doteq \varphi(\max(\text{supp}(r))) \in \omega$ for $r \neq 0$, we argue by induction on $n(a_1) + n(a_2)$. Let $w_i = \max(\text{supp}(a_i)) \in S$, for $i = 1,2$.
Since $a_1b_1 + a_2 b_2 = 0$, we can find $v_1, v_2 \in S$ such that $w_1 v_1 = w_2 v_2$. Swapping $a_1$ and $a_2$ if needed, we can assume that $w_1 \le w_2$, so that $w_1$ is an initial subword of $w_2$. Therefore we can find $s \in S$ and $\lambda \in F$ such that either $a_2' = 0$ or
$n(a_2') < n(a_2)$ where $a_2' = a_2 - \lambda a_1 s$. We set $a_1' = a_1$,
$b_1' = b_1 + \lambda sb_2$ and $b_2' = b_2$. Then we have $a_1' b_1' + a_2'b_2' = 0$ and $(a_1', a_2')$ is obtained from $(a_1, a_2)$ by right-multiplication of a matrix in $E_2(R)$. If $a_2' = 0$, we are done. Otherwise the induction hypothesis applies to $(a_1', a_2')$, which yields the result.

We are now in position to prove the claim.

*Proof of the claim.* Let $A = (a_{ij}) \in GL_2(R)$ with $R = F\langle x,y \rangle$. As $A$ has a right inverse, we can find $b_1, b_2 \in R$, not both zero, such that $a_{11} b_1 + a_{12}b_2 = 0$. By the above lemma, we can assume that $a_{12} = 0$. It follows that $a_{11} \in F^{\times}$, so that $A$ can be reduced to an invertible diagonal matrix through an elementary row operation. Therefore $A \in E_2(R)D_2(R)$.

[1] H. Bass, "Projective modules over free groups are free", 1964.

[2] P. M. Cohn, "On the structure of the $GL_2$ of a ring", 1966.