Definitions: A function $f \in C_0(\mathbb{R})$ is quasi-invertible if $1 \notin \operatorname{ran}(f)$. The quasi-inverse of such an $f$ is $\frac{f}{f-1}$.
Some discussion: Let $C_1(\mathbb{R})$ denote the continuous functions $\mathbb{R} \to \mathbb{C}$ which tend to $1$ at $\pm \infty$. Notice $f \mapsto 1-f$ is an involution of the set of all functions $\mathbb{R} \to \mathbb{C}$ which exchanges $C_0(\mathbb{R})$ and $C_1(\mathbb{R})$. Note $C_1(\mathbb{R})$ is closed under multiplication and contains the constant $1$ function. The elements of $C_0(\mathbb{R})$ corresponding (under the involution) to invertible elements of $C_1(\mathbb{R})$ are the quasi-invertible functions, and quasi-inversion corresponds to ordinary inversion $f \mapsto \frac{1}{f}$ in $C_1(\mathbb{R})$.
Question: Let $f \in C_0(\mathbb{R})$ be quasi-invertible (that is, $1 \notin \operatorname{ran}(f)$). If $f$ is in the range of the Fourier transform $L^1(\mathbb{R}) \to C_0(\mathbb{R})$, is its quasi-inverse $\frac{f}{f-1}$ also in the range of the Fourier transform?