Since $\mathcal{A}$ lies in the ring of holomorphic functions on $\Omega$ (I will call this ring $C^\omega(\Omega)$), it is important to understand what does $z^\alpha$ means for $\alpha \in \mathbb{C}$. The meaning is clear when we set $z^\alpha=e^{ log(z)\alpha}$, and similarly $(1-z)^\beta = e^{log(1-z)\beta}$. Clearly, because of the choice of the region $\Omega$, the function $log$ used above can be well-defined. This choice of a $log$ function makes us wish to understand what happens when we make a change to this choice.
$\mathcal{A}\subset C^\omega(\Omega)$ is a $\mathbb{C}$-vector space, with a generating set given by $\{z^{\alpha}(1-z)^\beta\}_{(\alpha,\beta)\in \mathbb{C}^2}$. It is certainly clear that this generating set is not linearly independent. For a while, assume that it is possible to define the following $\mathbb{C}$-linear functions $T_{x_i}:\mathcal{A}\rightarrow \mathcal{A},\ (i=0,1)$ such that the following happens
$$
T_{x_0}:z^\alpha(1-z)^\beta \mapsto z^\alpha(1-z)^\beta e^{(2\pi i)\alpha}
$$
$$
T_{x_1}:z^\alpha(1-z)^\beta \mapsto z^\alpha(1-z)^\beta e^{(2\pi i)\beta}
$$
Now, we should be worried if this linear map is well-defined. For instance, if an element is written in two different ways as a $\mathbb{C}$-linear combination of elements $\{z^{\alpha}(1-z)^\beta\}_{(\alpha,\beta)\in \mathbb{C}^2}$, would these operators defined on this generating set yield consistent output? I will postpone this concern to when I define the maps $T_{x_0}, T_{x_1}$ more geometrically after justifying their importance.
Let us first see the benefit that these operators will have. First of all, each element of our generating set $\{z^{\alpha}(1-z)^\beta\}_{(\alpha,\beta)\in \mathbb{C}^2}$ is an eigenvector of this linear map. Now note the following lemma.
Lemma: For a $\mathbb{C}$-vector space $V$ and a map $T:V\rightarrow V$, a finite collection of eigenvectors corresponding to distinct eigenvalues are linearly independent over $\mathbb{C}$.
Proof can be found here
Now suppose we have a set $F \subset \mathbb{C}^2$ which contains a linear dependency relation of the following type ($c_{\alpha,\beta} \in \mathbb{C}$, $G \subset F$ is a non-empty finite set)
$$
\sum_{(\alpha,\beta) \in G} c_{\alpha,\beta}\ z^\alpha (1-z)^\beta =0 \ \ \ \ \ ...(1)
$$
Without loss of generality, we assume that the choice of $G$ is such that $G$ contains the minimum possible number of terms. This leads us to believe that all the terms of the type $z^\alpha (1-z)^\beta$ in the above sum must reside in the same eigenspace of $T_{x_i},\ (i=0,1)$. Indeed, because if this were not the case, there would be elements of at least two different eigenspaces in the above sum and it should be possible to segregate the terms to get eigenvectors of different eigenspaces. Each of those individual sums will then have to be zero by the previous lemma causing a contradiction of the minimality of $G$.
Since the eigenvalues are the same, we assume that it is $e^{(2\pi i)a}$ for $T_{x_0}$ and $e^{(2\pi i)b}$ for $T_{x_1}$ for some $a,b \in \mathbb{C}$. This tells us that in the above sum, each $z^\alpha (1-z)^\beta$ is actually of the form $z^{a+m} (1-z)^{b+n}$ for some $m,n \in \mathbb{Z}$. We promptly divide the sum by $z^a (1-z)^b$ to modify our relation (1) and get the following
$$
\sum_{(m+a,n+b) \in G} c_{(m+a),(n+b)}\ z^m (1-z)^n = 0
$$
More importantly, $G$ is just a translation of some set $G' \in \mathbb{Z}^2$ via $(a,b) \in \mathbb{C}^2$. So with the above discussion, we arrive at the following lemma.
Lemma: For $F \subset \mathbb{C}^2$, $\{z^\alpha (1-z)^\beta \}_{(\alpha,\beta) \in F }$ can be linearly dependent over $\mathbb{C}$ (i.e. there is a finite subset of $\{z^\alpha (1-z)^\beta \}_{(\alpha,\beta) \in F }$ that is linearly dependent over $\mathbb{C}$) if and only if there exists a finite set $G' \subset \mathbb{Z}^2$ such that $\{ z^m (1-z)^n \}_{(m,n) \in G'}$ are linearly dependent over $\mathbb{C}$ and $G'$ can be translated to become a subset of $F$ (i.e. there exists $(a,b)\in \mathbb{C}^2$ such that $(m+a,n+b) \in F,\ \forall (m,n) \in G'$).
Proof: If $G'$ can be translated to fit inside $F$ then clearly we are done (multiplying our linear dependence relation inside $\{z^\alpha (1-z)^\beta \}_{(\alpha,\beta) \in G' }$ by a suitable $z^a(1-z)^b$ gives us a dependence relation in $\{z^\alpha (1-z)^\beta \}_{(\alpha,\beta) \in F }$ ). Moreover, by the previous discussion, all minimal linear dependencies in the set $\{z^\alpha (1-z)^\beta \}_{(\alpha,\beta) \in F }$ can only be of the above type.
We conclude our discussion with the following Proposition, which will give us a way to creating subsets $D \subset \mathbb{C}^2$ such that $\{z^\alpha (1-z)^\beta\}_{(\alpha,\beta)\in D}$ becomes a basis of $\mathcal{A}$.
Proposition: Suppose $A\subset \mathbb{Z}^2$ is such that $\{z^m (1-z)^n\}_{(m,n)\in A}$ is a basis of the vector space $\mathbb{C}\{z^m(1-z)^n\}_{(m,n)\in \mathbb{Z}^2} \subset C^\omega(\Omega)$, then $\{z^\alpha (1-z)^\beta\}_{(\alpha,\beta)\in D}$ is a basis of the vector space $\mathbb{C}\{z^\alpha(1-z)^\beta\}_{(\alpha,\beta)\in \mathbb{C}^2}$ where $D\subset \mathbb{C}^2$ is given by
$$
D = \{ (m+a,n+b)|\ (m,n) \in A,\ a,b \in \mathbb{C},\ 0\leq \Re(a) < 1,\ 0\leq \Re(b) < 1 \}
$$
Proof: Let us denote the eigenspace of $T_{x_i}$ corresponding to the eigenvalue $e^{(2\pi i )\alpha}$ by $V^i_{\alpha}$ ($i=0,1$). Let $W_{\alpha,\beta} = V_\alpha^0 \cap V_\beta^1 $. Clearly, $W_{0,0}\cap \mathcal{A} = \mathbb{C}\{z^m(1-z)^n\}_{(m,n)\in \mathbb{Z}^2}$. All elements of $W_{0,0}\cap\mathcal{A}$ can therefore be generated by $\{z^\alpha (1-z)^\beta\}_{(\alpha,\beta)\in D}$. Similarly, for $a,b \in \mathbb{C},\ 0\leq \Re(a) < 1,\ 0\leq \Re(b) < 1$ all elements of $W_{a,b}\cap \mathcal{A}$ can be generated by $\{z^{(m+a)} (1-z)^{(n+b)}\}_{(m,n)\in A}$. This way, we see that $\{z^\alpha (1-z)^\beta\}_{(\alpha,\beta)\in D}$ is a generating set of $\mathcal{A}$.
The fact that it is a basis can be established by the previous lemma. If D had a linearly dependence inside it, then we should be able to find a minimal linearly dependent set within some $\{z^\alpha (1-z)^\beta\}_{(\alpha,\beta)\in D} \cap W_{a,b}$ for some $a,b \in \mathbb{C}$. But, this set cannot contain any linear dependence by the choice of A.
One such $D$ can be constructed using $A=\{(-m,0),m\in \mathbb{N}\cup\{0\}\}\cup (\{0\}\times \mathbb{Z})$.
So what are these operators $T_{x_0}, T_{x_1}$?
In what follows, I will establish how the mysterious operators $T_{x_i}:\mathcal{A}\rightarrow \mathcal{A}$ emerge. By the virtue of the construction, it will become clear that the operators are well-defined linear functions on $\mathcal{A}$
We define $B=\mathbb{C}\backslash \{0,1\}$. Thus, we have a inclusion $\Omega \hookrightarrow B$. Let $p: E\rightarrow B$ be the universal covering space of $B$ such that $e \in E$ is mapped to $p(e)=b \in \Omega \subset B$. Since $\Omega$ is simply connected, the inclusion $\Omega \hookrightarrow B$ can be lifted to $E$ such that $\Omega \hookrightarrow E$ maps $b$ to $e$.
$$\require{AMScd} \begin{CD} \Omega @>\text{inclusion}>> E\\ @| @VVV \\ \Omega @>\text{inclusion}>> B \end{CD}$$
It is well known that $E$ is also a Complex $1$-manifold (a Riemann surface) like $\Omega$ and $B$ and it is possible to define holomorphic functions of $E$. Moreover, we have a natural action of the fundamental group $\pi_1(B,b)$ on $E$ as holomorphic deck transformations ( each $g \in \pi_1(B,b)$induces a biholomorphic homeomorphism $t_g:E\rightarrow E$ such that $p \circ t_g = p$ ). Hence we have the following diagram
$$\require{AMScd} \begin{CD} \Omega @>\text{inclusion}>> E @>t_g>> E\\ @| @VpVV @VpVV \\ \Omega @>\text{inclusion}>> B @= B\end{CD} $$
It is also well-known that the $\pi_1(B,b)$ is the free group generated by two generators. We choose two generators $x_0, x_1 \in \pi_1(B,b)$ that correspond to the path homotopy class of paths with endpoints at $b\in B$ and going around $0, 1$ in clockwise and anti-clockwise fashion respectively.
Corresponding to the diagram above, we have the opposite diagram induced on the ring of holomorphic functions on each of the Riemann surfaces
$$\require{AMScd} \begin{CD} C^\omega(\Omega) @<\text{Res}^{E}_{\Omega}<< C^\omega(E) @<T_g<< C^\omega(E) \\ @| @Ap_*AA @Ap_*AA \\ C^\omega(\Omega) @<\text{Res}^{B}_{\Omega}<< C^\omega(B) @= C^\omega(B)\end{CD} $$.
It is important to note that since $\Omega$ is embedded as open subset of $E$, the restriction map $\text{Res}^E_\Omega$ is injective. Morevoer, $\mathcal{A} \subset \text{Res}^E_\Omega(C^\omega(E))$. We can embed $\mathcal{A}$ into $C^\omega(E)$ by the following way. The following inclusion when right-composed to the restriction will give an identity map on $\mathcal{A}$
$$
\mathcal{A} \hookrightarrow C^\omega(E)
$$
$$
z^\alpha (1-z)^\beta \mapsto e^{\alpha \mathbf{log_0}} e^{\beta \mathbf{log_1}}
$$
Here $\mathbf{log_0}:E\rightarrow \mathbb{C}$ and $\mathbf{log_1}:E\rightarrow \mathbb{C}$ are holomorphic functions such that $\text{Res}^{E}_{\Omega}(\mathbf{log_0})(z) = log(z)$ and $\text{Res}^{E}_{\Omega}(\mathbf{log_1})(z) = log(1-z)$, for $z\in \mathbb{C}$ and the $log$ function is what was selected originally to define $\mathcal{A}$. Of course, we need to show that $\mathbf{log_0}, \mathbf{log_1}$ must exist. This detail can be given on request.
With the setup above, $T_{x_0}$ and $T_{x_1}$ have a clear meaning as linear functions on $C^\omega(\Omega)$ when we put $g=x_0,x_1$. For the purpose of the above discussion, we will have to restrict these functions to $\mathcal{A}$ where the inclusion $\mathcal{A} \hookrightarrow C^\omega(E)$ is as defined above. It can be easily checked that they give the output as desired. For instance, $T_{x_0}(e^{\alpha \mathbf{log_0}} e^{\beta \mathbf{log_1}} )(z) =e^{\alpha (\mathbf{log_0}(z)+2\pi i)} e^{\beta \mathbf{log_1}(z)}$, for $z \in E$ and the output is what is needed in the above discussion follows.
Hence with this, I finish my answer. Please let me know if there is a need for greater clarity anywhere in particular (especially in the places where I use the word "well-known").
