Let's consider $B(x) = A^\alpha(x)$. (My understanding of OP's question is ultimately considering the case where $\alpha=1/2$ and expressing $B_i$ in terms of $A_i$.) If $B(x) = A^\alpha(x)$, then
\begin{align*}
B_k &= \frac{1}{k A_{0}} \sum_{i=1}^{k} (i(\alpha+1) - k) A_i B_{k-i}\\
\text{and }B_0 &= A_0^\alpha.
\end{align*}
Proof: Since $\alpha$ and $a$ are a little close to each other, we're going to deviate slightly from OP's notation and refer to $A(x) = \sum A_i x^i$ and $B(x) = \sum B_i x^i$.
\begin{align*}
B(x) &= A^\alpha(x) \\
B'(x) &= \alpha A^{\alpha-1}(x)A'(x). \\
A(x)B'(x) &= \alpha A^{\alpha}(x)A'(x) \\
&= \alpha A'(x) B(x) \\
\end{align*}
Now let's look at the terms of degree $k-1$ on each side of $A(x)B'(x) = \alpha A'(x) B(x)$.
\begin{align*}
\require{color}
\sum_{j=0}^{\color{red}{k-1}} (j+1) A_{k-j-1} B_{j+1} &= \sum_{i=0}^{\color{red}{k-1}} \alpha (i+1) A_{i+1} B_{k-i-1} \\
\color{red}{k A_{0} B_{k} } \color{blue}{ + \sum_{j=0}^{k-2} (j+1) A_{k-j-1} B_{j+1}} &= \color{red}{\alpha k A_{k} B_{0}} \color{black}{+} \sum_{i=0}^{k-2} \alpha (i+1) A_{i+1} B_{k-i-1} \\
k A_{0} B_{k} &= \alpha k A_{k} B_{0} + \sum_{i=0}^{k-2} (i+1) \alpha A_{i+1} B_{k-i-1} \color{blue}{- \sum_{j=0}^{k-2} (j+1) A_{k-j-1} B_{j+1} }\\
\end{align*}
If we let $i = k - 2 - j$, basically just reversing the order we do this sum, we see that $\sum_{j=0}^{k-2} (j+1) A_{k-j-1} B_{j+1} = \sum_{i=0}^{k-2} (k-i-1) A_{i+1} B_{k-i-1}$. And so,
\begin{align*}
k A_{0} B_{k} &= \alpha k A_{k} B_{0} + \sum_{i=0}^{k-2} \alpha(i+1) A_{i+1} B_{k-i-1} - (k-i-1) A_{i+1} B_{k-i-1} \\
&= \alpha k A_{k} B_{0} + \sum_{i=0}^{k-2} (\alpha(i + 1) - k + (i + 1)) A_{i+1} B_{k-i-1} \\
k A_{0} B_{k} &= \alpha k A_{k} B_{0} + k \sum_{i=0}^{k-2} ((\alpha+1)(i + 1) - k) A_{i+1} B_{k-i-1} \\
B_{k} &= \alpha A_{k} \frac{B_{0}}{A_{0}} + \frac{1}{k A_{0}} \sum_{i=0}^{k-2} ((\alpha+1)(i + 1) - k) A_{i+1} B_{k-i-1} \\
\end{align*}
And we can make this a little tighter by messing with the index on the summation:
\begin{align*}
B_{k} &= \color{red}{\alpha A_{k} \frac{B_{0}}{A_{0}}} \color{black}{+} \frac{1}{k A_{0}} \sum_{i=1}^{k-1} (i(\alpha+1) - k) A_i B_{k-i}\\
&= \frac{1}{k A_{0}} \sum_{i=1}^{\color{red}{k}} (i(\alpha+1) - k) A_i B_{k-i}\\
\end{align*}
If we go back to $B(x) = A^\alpha(x)$ and set $x=0$ on both sides, that gives us $B_0 = A_0^\alpha$. $\tag*{$\blacksquare$}$
This argument is from Donald Knuth's The Art of Computer Programming v2 ch4.7, where he attributed it to Leonard Euler's Introductio in Analysin Infinitorum. He assumed that $A_0 = 1$ and subsequently $B_0 = 1$.
Now, if you're like me when I first read this argument, you might find this all a little suspicious. If $A(x)=1+5x+10x^2+10x^3+5x^4+x^5$ and $\alpha=1/5$, then we can get that $(1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5)^{1/5} = 1 + x$ by just computing $B_0$ and $B_1$ and in the process only use the information in $A_0$ and $A_1$. This may seem true, but it is not. To see that, keep in mind that these computations are treating these polynomials like power series. To show that $A^{1/5}(x)=1+x$, we need to also show that $B_i=0$ for all $i>1$, which isn't hard, but may not be immediately obvious. For $i \in [2,6]$, a lot of nice cancelation happens. For $i>6$, the combination of $A_i=0$ and $B_j=0$ for $j \in [2,i-1]$ make it easy to argue that $A_i=0$ for all $i>6$ as well. The short of it is that you really do need all the information in $A(x)$ to argue that $B(x)$ has finite degree, that is that it actually is a polynomial. That's all peripheral to OP's question, but I get why this might be somewhat controversial since this is a closed form answer that doesn't have that illuminating shine.
