I was overwhelmed by the number of responses I got (10 answers so far!). Of course, all of them got my upvote. This was fun, thanks guys for your thoughts, comments, etc. I know that by now most of you know what the flaw is, at least the one I meant. People express things differently, and there's always room for misunderstandings, so I will try to clearly formulate what I think is the most important flaw in that derivation. I'm aware of the fact that not everybody will agree and that's fine. I'm happy to be able to discuss this sort of esoteric DSP topics with such sharp minds as y'all are! Here we go.
My first claim is that each and every equation in my question is correct. However, the derivation and motivation of some of them is totally wrong and misleading, and that "derivation" can only exist because the author knew what the result was supposed to look like.
Eq. (3) in the question ($f[n]-f[n-1]=\delta[n]$) is correct for the given sequence $f[n]$ (Eq. $(2)$ in the question), but it is clearly also correct for all sequences of the form $$f[n]=u[n]+c\tag{1}$$ with some arbitrary constant $c$. So, according to the derivation, the resulting DTFT $F(\omega)$ should be the DTFT of all sequences of the form $(1)$, regardless of the value of the constant $c$. That's of course non-sense because the DTFT is unique. Specifically, using that very "proof" I could "show" that $F(\omega)$ as given in Eq. $(5)$ of my question (or Eq. $(3)$ below) is actually the DTFT of $u[n]$ that we're looking for. So why bother splitting up $u[n]$ as in Eq. $(1)$ of the question?
However, it is true that the DTFTs of all sequences $(1)$ do satisfy Eq. $(4)$ in the question (repeated here for convenience): $$F(\omega)\left(1-e^{-j\omega}\right)=1\tag{2}$$ But now comes the actual mathematical flaw: From Eq. $(2)$ it is incorrect to conclude $$F(\omega)=\frac{1}{1-e^{-j\omega}}\tag{3}$$ Eq. $(3)$ is only one of infinitely many possible solutions of $(2)$, and it conveniently happens to be the one needed by the author to arrive at the correct end result. Eq. $(3)$ is the DTFT of $f[n]$ in $(1)$ with $c=-\frac12$, but from the given derivation there is no way to know that.
So how can we avoid that mathematical error and use $(2)$ to derive the DTFTs of $all$ sequences $(1)$, with any constant $c$? The correct conclusion from $(2)$ is $$F(\omega)=\frac{1}{1-e^{-j\omega}}+\alpha\delta(\omega)\tag{4}$$ with some yet undetermined constant $\alpha$. Plugging $(4)$ into the left-hand side of $(2)$ gives $$1+\alpha (1-e^{-j\omega})\delta(\omega)=1+\alpha (1-e^{-j\omega})\Big{|}_{\omega=0}\cdot\delta(\omega)=1+0\cdot\delta(\omega)=1$$ So all functions $F(\omega)$ given by $(4)$ satisfy $(2)$, as required.
The constant $\alpha$ in $(4)$ can be determined from the value of $f[n]$ at $n=0$: $$f[0]=1+c=\frac{1}{2\pi}\int_{-\pi}^{\pi}F(\omega)d\omega=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{d\omega}{1-e^{-j\omega}}+\frac{\alpha}{2\pi}\tag{6}$$ It can be shown, and also WolframAlpha agrees, that the Cauchy principal value of the integral in $(6)$ is $$PV\int_{-\pi}^{\pi}\frac{d\omega}{1-e^{-j\omega}}=\pi\tag{7}$$ From $(6)$ and $(7)$ we get $$\alpha=\pi (1+2c)\tag{8}$$ So for $c=-\frac12$ we get $\alpha=0$ (which corresponds to the original sequence $f[n]$ as used by the author of the proof), and for $c=0$ (i.e., for $f[n]=u[n]$) we have $\alpha=\pi$, which finally gives us the desired DTFT of $u[n]$: $$U(\omega)=\frac{1}{1-e^{-j\omega}}+\pi\delta(\omega)\tag{9}$$