No. Let $k$ be a field of characteristic $0$. Consider the symmetric algebra on one generator $k[x]$, with comultiplication $x \mapsto x\otimes 1 + 1\otimes x$. It has a Hopf pairing with itself, given by $\langle x^m,x^n\rangle = n! \hspace{.5ex} \delta_{m=n}$. Then consider the bases of $k[x]$ given by expansion around $1$, i.e. $e_n = (x-1)^n$. The dual basis, if it exists, includes $e^0$ such that $\langle e^0, (x-1)^n\rangle = \delta_{0,n}$. So suppose that $e^0 = \sum a_n x^n$; then:
$$ \begin{aligned}
1 & = a_0 \\
0 & = a_1 - a_0 \\
0 & = a_2 - 2a_1 + a_0 \\
\dots & \phantom= \dots \\
0 & = \sum_{k=0}^n (-1)^{n-k} \hspace{.5ex} n!\hspace{.5ex} a_n \hspace{.5ex} \binom{n}{k} \\
\dots & \phantom= \dots \\
\end{aligned} $$
The solution is that all $a_n = 1/n!$, i.e. $e^0 = \sum x^n/n! = \exp(x)$. But this is not a polynomial.

In the case you ask about, you will similarly have some bases with dual bases, and some without.