The sentence "we may as well" is unclear: $f$ is given, so $f(0)$ is what it is. I suspect that you are working on the proof of some theorem, and you want to add an assumption that is not among those of the theorem itself. In this case, in my opinion, it is enough to remark that $f-f(0)$ is a $C^1$ function, with the same period as $f$, which vanishes at $0$. Moreover, being $f$ of class $C^1$, its Fourier series converges uniformly and hence pointwise to $f$. As a consequence, up to replacing $f$ with $f-f(0)$, you can think that $f(0)=0$ and that the Fourier series of $f$ converges to $f(0)=0$ at $x=0$.
Moving to your second question, the equation $e^{ix}=1$ is solved in $[0,2\pi)$ only by $x=0$. Since the map $x \mapsto e^{ix}-1$ is continuous (here you need to know something about vector-valued functions), the function $g$ is continuous away from $x=0$. You need to check that $g$ is also continuous at $x=0$. Now, $ e^{ix}-1 = \sum_{n=1}^\infty \frac{(ix)^n}{n!}, $ and $ g(x) = \frac{f(x)}{\sum_{n=1}^\infty \frac{(ix)^n}{n!}} = \frac{f(x)}{x \left( i+\sum_{n=2}^\infty \frac{(ix)^n}{n!} \right)} = \frac{f(x)}{x} \frac{1}{i+\sum_{n=2}^\infty \frac{(ix)^n}{n!}} = \frac{f'(0)}{i}+o(1) $ as $x \to 0$. We have proved that the function $ \tilde{g}(x)=\begin{cases} \frac{f(x)}{e^{ix}-1} &\text{if $x \neq 0$} \\ \frac{f'(0)}{i} &\text{if $x=0$} \end{cases} $ is continuous on $[0,2\pi)$, which is a more precise statement than the one you wrote.