= Solution
Write $X=I+E$. Use a submultiplicative <Sobolev algebra> <norm> for which the Fourier projections are contractive, and its compatible <matrix>/column <norms>. The numeric hypothesis $\|E\|<1$ is understood in this algebra <norm>; the unscaled Fourier <Sobolev norm> need not have multiplication constant one.
Let $A_{>0}$ and $A_{<0}$ denote the strictly positive and strictly negative Fourier subspaces. The equation $y_+=-P_{>0}(E+Ey_+)$ is a contraction on $M_n(A_{>0})$, so it has a unique solution. Put $Y_+=I+y_+$ and $Z_-=XY_+$. The equation says $Z_-\in M_n(A_-)$, and $Y_+(0)=I$. Similarly the contraction $v_-=-P_{<0}(E+v_-E)$ gives $V_-=I+v_-$ with $Z_+=V_-X\in M_n(A_+)$. No invertibility of these solutions has yet been assumed.
Now $V_-Z_-=Z_+Y_+$ belongs to both Fourier subalgebras, hence is a constant <matrix> $C$. Taking constant coefficients on either side shows $C=Z_-(0)=Z_+(0)$. We claim $C$ is invertible. If $Cv=0$, then the Hardy compression $T_X=P_{\geq0}X$ annihilates $Y_+v$, because $P_{\geq0}Z_-v=Cv=0$. But $T_X=I+P_{\geq0}E$ on the Hardy <Sobolev space> has inverse given by a <Neumann series>, since the perturbation has <norm> less than one. Thus $Y_+v=0$, and its constant coefficient is $v$, forcing $v=0$. Finite-dimensionality proves the claim.
Consequently $C^{-1}Z_+$ is a left inverse of $Y_+$, and $C^{-1}V_-$ is a left inverse of $Z_-$. <Matrix> entries are functions in a commutative algebra, so pointwise finite-dimensional inversion makes these left inverses two-sided inverses in their respective Fourier subalgebras. Set
$$
\boxed{X_-=Z_-,\qquad X_+=Y_+^{-1}.}
$$
Then $X=X_-X_+$, both factors are invertible in the stated algebras, and $X_+(0)=I$. If two normalized factorizations exist, $\widetilde X_-^{-1}X_-=\widetilde X_+X_+^{-1}$ lies in $A_-\cap A_+$ and is constant; evaluating the analytic side at zero makes it $I$. This proves uniqueness. The argument is <near-identity Sobolev loop factorization> and, crucially, proves invertibility rather than treating the contraction solution itself as automatically invertible.
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