discrete signals - Where is the flaw in this derivation of the DTFT of the unit step sequence $u[n]$?

This question is related to this other question of mine where I ask for derivations of the discrete-time Fourier transform (DTFT) of the unit step sequence $u[n]$. During my search for derivations I found one which is amazingly simple. I first saw it on page 138 of this book by B.A. Shenoi. I also came across it on mathematics.SE in this answer.

Since the argument is short and simple I will repeat it here for convenience.

The unit step sequence can be written as

$$u[n]=f[n]+\frac12\tag{1}$$ with $$f[n]=\begin{cases}\frac12,\quad n\ge 0\\-\frac12,\quad n<0\end{cases}\tag{2}$$ Obviously, $$f[n]-f[n-1]=\delta[n]\tag{3}$$ Applying the DTFT on both sided of $(3)$ gives $$F(\omega)\left(1-e^{-j\omega}\right)=1\tag{4}$$ where $F(\omega)$ is the DTFT of $f[n]$. From $(4)$ we get $$F(\omega)=\frac{1}{1-e^{-j\omega}}\tag{5}$$ From $(5)$ and $(1)$ we get for the DTFT of $u[n]$ $$U(\omega)=F(\omega)+\pi\delta(\omega)=\frac{1}{1-e^{-j\omega}}+\pi\delta(\omega),\quad -\pi\le\omega <\pi\tag{6}$$ where I've used $\text{DTFT}\{1\}=2\pi\delta(\omega)$, $-\pi\le\omega <\pi$.

Eq. $(6)$ for the DTFT of $u[n]$ is no doubt correct. However, the derivation is flawed.

The question is: find and explain the flaw in above derivation.

Please prepend your answer with the spoiler tag >!.

Answer

There are infinitely many signals that make the following equality hold:

$$y[n]-y[n-1]=\delta[n] \qquad (1)$$ The only thing that matters is that $y[0]-y[-1]=1$, and then the rest of the coefficients of $y$ can be determined under the restriction that Eq. $(1)$ states (i.e. the substraction of consecutive samples must be $0$ for $n\neq 0$). In other words, Eq. $(1)$ will be achieved by any signal $y[n]$ such that $$y[0]=y[-1]+1 \land y[n]=y[n-1] \ \forall n\neq0$$ Another way to see this is that any function that is basically $u[n]$ with an offset (a constant value added) will satisfy $(1)$. This explains the statement made by robert bristow-johnson in his answer: differentiators destroy this information (such as taking a derivative in continuous time destroys evidence of any constant value in the original function).

To sum up, I believe that the proof is flawed because the procedure followed could use any function of the form $u[n]+C$ with $C\in\mathbb{R}$, and this would lead to many functions having the same Fourier transform, which is indeed wrong as the Fourier transform is a bijection. Maybe the author deliberately decided to ignore anything related to DC values, conscious that in order to show that $F(\omega)$ is the DTFT of $f[n]$ he would need the accumulation property (whose most popular proof is derived from the DTFT of the unit step - ergo, a pretty circular proof). The proof is not strictly wrong, as everything it states (the formulae for $F(\omega)$ and $U(\omega)$, the decomposition of the unit step, the difference equation) is true, but it would require the accumulation property to show why $F(\omega)$ doesn't have any Dirac deltas.