= Solution
{c}
Take $D=\hat R$. We first check that this <noncommutative formal power series ring> is a <noncommutative domain>. For nonzero $F,G\in D$, let $p,q$ be the smallest lengths of <words> with nonzero coefficients in $F,G$, respectively. Their degree-$p$ and degree-$q$ parts $F_p,G_q$ are nonzero <homogeneous polynomials> in the <free associative algebra>.
Choose <words> $u,v$ with nonzero coefficients in $F_p,G_q$. In $F_pG_q$, the coefficient of $uv$ is the product of those two coefficients: a <word> of length $p+q$ has a unique splitting into a prefix of length $p$ and a suffix of length $q$. This product is nonzero in $\mathbb Q$. The degree-$p+q$ part of $FG$ is exactly $F_pG_q$, since all lower parts of $F$ and $G$ vanish. Thus $FG\ne0$, proving that $D$ is a <noncommutative domain>.
By the defining property of the <free associative algebra>, there is a <ring homomorphism>
$$
f:R\longrightarrow D,\qquad f(x_1)=1+x_1,\qquad f(x_2)=2+x_2.
$$
The same substitution defines a <ring automorphism> of $R$, with inverse $x_1\mapsto x_1-1$, $x_2\mapsto x_2-2$. Composing this <ring automorphism> with the natural inclusion $R\hookrightarrow D$ proves that $f$ is <injective>.
Finally,
$$
\begin{aligned}
f(x_1+x_2)&=3+x_1+x_2,& f(x_1-x_2)&=-1+x_1-x_2,\\
f(x_1)&=1+x_1,&f(x_2)&=2+x_2.
\end{aligned}
$$
Their <constant terms> are $3,-1,1,2$, respectively, all nonzero. By the <unit criterion for noncommutative formal power series> proved in <2/c>, all four elements are <units>. Therefore $D$ and $f$ have every required property. The embedding changes the images of the generators, so the failure of the original monomial <left Ore set> in <2/b> presents no obstruction.
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