This question is a bit more subtle than I had originally thought (in the comments), but anyway here's an argument that seems to work. I will assume for notational convenience that the ultrafilter is on the first infinite ordinal $\omega$.
Fix $k \in \omega$ with $1/k < a$. The main claim is that any conull subset of $2^\omega$ with respect to the $(a, 1-a)$-product measure $\mu$ contains elements $x_0, \ldots, x_{k-1}$ such that $\bigcap_{i < k} x_i = \emptyset$ (that is, for each $n \in \omega$ there is $i < k$ with $x_i(n) = 0$). The trick is to make the situation "continuous" by introducing the function $f: [0,1)^\omega \to 2^\omega$ given by $f(y)(n) = 0$ if $y(n) < a$ and $f(y)(n) = 1$ if $y(n) \geq a$. It should be straightforward to check that if $\nu$ is the usual Lebesgue product measure on $[0,1)^\omega$, then $\nu(f^{-1}(A)) = \mu(A)$ for basic open (and thus all measurable) sets $A \subseteq 2^\omega$.
So suppose $A \subseteq 2^\omega$ is $\mu$-conull, thus $B = f^{-1}(A)$ is $\nu$-conull. Consider the $\nu$-preserving automorphism of simultaneous rotation by $1/k$, i.e., $g_k: [0,1]^\omega 0,1)^\omega \to [0,1)^\omega$ given by $g_k(y)(n) = y(n) + 1/k \1/k$ (mod 1$. $1$). Since $B$ is $\nu$-conull, there is some point $y \in \bigcap_{i < k} g_k^{-i}(B)$. Then the points $x_i = f(g_k^i(y))$ are as desired, since for any $y \in [0,1)^\omega$ and any $n \in \omega$, at least one of $g_k^i(y)(n) = y(n) + i/k \i/k$ (mod 1$ $1$) is less than $a$.
In particular, any $\mu$-conull set closed under finite intersection contains the empty set, so it can't be an ultrafilter.
