How far can we push the Fundamental Theorem of Calculus for Riemann integral?

Yes, if $f'$ is Riemann integrable then $\int_a^bf'=f(b)-f(a)$. I've posted a proof of this before, but it's simple enough that giving the proof again seems easier than trying to find that post:

Say $a=x_0<\dots<x_n=b$ is a partition of $[a,b]$. The Mean Value Theorem shows that there exists $\xi_j\in(x_{j-1},x_j)$ such that $$f(x_j)-f(x_{j-1})=f'(\xi_j)(x_j-x_{j-1}).$$So $$f(b)-f(a)=\sum_j(f(x_j)-f(x_{j-1}))=\sum_jf'(\xi_j)(x_j-x_{j-1}).$$But that last sum is precisely a Riemann sum for $\int_a^b f'$, so for any $\epsilon>0$ the last sum above is within $\epsilon$ of $\int_a^bf'$ if $\max_j(x_j-x_{j-1})$ is small enough.

So $$\left|f(b)-f(a)-\int_a^b f'\right|<\epsilon$$for every $\epsilon>0$.

Now what if $f$ is just differentiable on $[a,b]\setminus S$? No if we assume just that $S$ is a null set. I don't know the answer if $S$ is countable, but I suspect it's no. Yes if $S$ is finite (and $f$ is globally continuous):

Say $S=(a_j)$, where $a_1<\dots<a_n$. The case proved above shows that $$f(a_{j+1})-f(a_j)=\lim_{\epsilon\to0}(f(a_{j+1}-\epsilon)-f(a_j+\epsilon))=\lim_{\epsilon\to0}\int_{a_j+\epsilon}^{a_{j+1}-\epsilon}f'=\int_{a_j}^{a_{j+1}}f';$$now take the sum over $j$.