You forgot to check $F$ is well-defined since it is apparently dependent on the path you choose. First of all cover $\Omega$ by open balls. We shall show there exists a primitive on each ball.
Let $B(z_0,R)$ be a ball. For $z\in B(z_0,R)$, choose the radial path from $z_0$ to $z$, call that $\gamma_z$.
Next define $$F(z)=\int_{\gamma_z}f(\xi)d\xi$$
Then we observe that for $h$ small, $$\frac{1}{h}[F(z+h)-F(z)]=\frac{1}{h}\int_{L(z,z+h)}f(\xi)d\xi$$by Gousrat Theorem where $L(z,z+h)$ is the straight line joining $z$ to $z+h$
Then we get $$\frac{1}{h}[F(z+h)-F(z)]=\frac{1}{h}\int_0^1f(z+\theta h)hd\theta=\int_0^1f(z+\theta h)d\theta \rightarrow f(z)$$as $h\rightarrow 0$ by your favourite convergence theorem.
Thus we have showed the existence of an anti-derivative on an open ball.
This in particular shows that integral of a holomorphic function on a closed curve in any open ball is 0.
Now let $H$ be a fixed end point homotopy between two paths $\gamma_0,\gamma_1$ in a region $\Omega$
Say $H: I^2\rightarrow \Omega$
Choose a partition of $I^2$ into a grid $\{G_{ij}\}$ so that any small tile $G_{ij}$ falls into an open ball in $\Omega$ via $H$ using continuity and compactness. Join the corners of the tiles in $\Omega$ by straight lines and for the sake of simplicity call them $G_{ij}$ as well.
Then one can write $$\int_{\gamma_0}f(\xi)d\xi -\int_{\gamma_1}f(\xi)d\xi =\sum_{i,j}\int_{\partial G_{ij}} f(\xi) d\xi $$
Each term in the last sum is $0$ as it is the integral of a holomorphic function on an open ball by our choice of the partition.
This shows integral of a holomorphic function on 2 fixed end-point homotopic curves is the same. In particular this shows integral of a holomorphic function on a closed curve in any simply-connected domain is 0.
Then you can proceed as you were doing.
If you are interested in a purely algebraic topology approach, here's one way to proceed.
We have solved the primitive problem locally an open cover say balls $\mathcal B=\{ B_i\}$. Any $2$ such solutions on a ball differ by a constant. Say we fix a local solution $\{f_i \}_i$ on the local cover $B_i$
Then on the intersection $B_i\cap B_j$ we get a complex number $c_{ij}$ such $f_i-f_j=c_{ij}$. Thus we get a co-cycle in the Cech cohomology group $\hat {H^1}(\Omega ;\mathcal B)$
The cover we chose was a Leray cover as intersections of convex sets are convex and hence all contractible. So the obstruction is an element of $\hat{H^1}(\Omega; \mathbb C)\cong {H_{dR}^1}(\Omega; \mathbb C)$
For simply-connected smooth manifolds, the $1$st de-Rham cohomology group is $0$ and hence we get the obstruction we got isn't an obstruction at all.
(Without making use of the assumption quoted in the question)$f$ has a Taylor expansion at $z_0$ with radius of convergence $r\in\mathopen]0\mathbin;+\infty]$.
Without loss of generality we may also suppose that $f(z_0)=0$ (replacing $f$ by $f-f(z_0)$).
That is $f(z)=\sum_{k=1}^{+\infty} a_k(z-z_0)^k$.
If all $a_k$ were 0, $f$ would be constant; hence $\{k\in\Bbb{N} \mid a_k\neq0 \}\neq\emptyset$, and this set has a least element $n$.
Now $f(z)=(z-z_0)^n\sum_{k=n}^{+\infty} a_k(z-z_0)^{k-n}
=\color{blue}{(z-z_0)^n g(z-z_0)}$, where $g(z)=\sum_{k=0}^{+\infty}a_{k+n}\,z^k$, and the radius of convergence of $g$ is also equal to $r>0$.
Since $g(0)=a_n\neq0$ and $g$ is continuous at $0$ there exists some $\varrho\in\mathopen]0\mathbin;r\mathclose[$ such that $g(z)\neq0$ for any $z\in D(0\mathbin;\varrho)$.
On the disk $D(z_0\mathbin;\varrho)$, $f$ has no other zero than $z_0$.
Best Answer
For (b) consider $\Omega = \Bbb C \setminus \{y i\mid y\ge 0\}$ and $f(z) = \log(iz)- \frac{\pi}{2}i$.