You can consider the set of probabilities over $\mathcal{F}$ as a subset of the linear space $V(\mathcal{F})$ of all finitely additive (bounded) scalar measures over $\mathcal{F}$ endowed with the variation norm (see Theory of charges (K. Bhaskara Rao, M. Bhaskara Rao), chapter 7). We need to show that this space is the topological dual of another (locally convex) topological vector space.
When $\mathcal{F}$ is a Boolean algebra of subsets of $\Omega$ (which in this case does not have to be a topological space) we define $S(\mathcal{F})$ as the linear space generated by $\{\chi_A:\ A\in\mathcal{F}\}$ (the characteristic functions of the sets in $\mathcal{F}$). $S(\mathcal{F})$ is called the space of simple functions.
Now, for every $f\in S(\mathcal{F})$, $\Vert f\Vert_s:=\sup\vert f\vert<\infty$ so $(S(\mathcal{F}),\Vert\cdot\Vert_s)$ is a normed space. It is not hard to prove that the dual of $(S(\mathcal{F}),\Vert\cdot\Vert_s)$ is the space $V(\mathcal{F})$. In fact, on one hand every $\lambda\in V(\mathcal{F})$ defines the bounded linear functional $f\mapsto \int f\ d\lambda$ (for $f\in S(\mathcal{F})$), on the other, every $x^*\in S(\mathcal{F})^*$ defines on $\mathcal{F}$ the measure $\lambda_{x^*}(A):=x^*(\chi_A)$ (for $A\in\mathcal{F}$). See Topological Riesz spaces and measure theory (Fremlin) for a more complete reference.
Having showed that the set of probabilities over $\mathcal{F}$ is contained in the topological dual of the normed space $S(\mathcal{F})$, it is clear why we can talk about weak$^*$-convergence.
PS: It is worth observing that by the Stone representation Theorem for Boolean rings any Boolean ring $\mathcal{R}$ is isomorphic to the ring of clopen sets in a locally compact Hausdorff space (see Measure theory Vol III (Fremlin), or this survey by Tao). Following this line, the approach showed by Tomasz can be proved to be much closer to the one exposed so far than someone would think. It would be interesting to go through this idea.
Best Answer
No. Take $\Omega=\mathbb N$ and $\Sigma$ the power set. As wikipedia says, then $ba(\Sigma)=ba=(\ell^\infty)^*$. However, the collection of probability measures is just the collection of $(x_n)\in\ell^1$ (as a measures are countably additive) with $x_n\geq 0$ for all $n$, and $\sum_n x_n=1$. This is not weak$^*$-closed in $(\ell^\infty)^*$. For example, any limit point of the set $\{\delta_n:n\in\mathbb N\}$, where $\delta_n\in\ell^1$ is the point mass at $n$, is a member of $ba \setminus \ell^1$.