If $G$ is a semigroup and $\beta G$ is the set of ultrafilters on $G$, then $\beta G$ is also a semigroup: given $p,q\in \beta G$, we define $$p*q := \{E\subseteq G : \{g\in G : g^{-1}A\in q\}\in p\}.$$ This is a strange definition at first, but indeed it is an associative binary operation on $\beta G$, and we have an embedding of semigroups $G\hookrightarrow \beta G$ by identifying each element of $G$ with its point-mass.
An ultrafilter $p$ can be identified with its indicator function $\chi_p:\mathcal{P}(G)\rightarrow \{0,1\}$, where $\mathcal{P}(G)$ is the power set of $G$. This is a finitely-additive probability measure on $G$.
Thus let $\mathrm{Pr}^0(G)$ be the space of all finitely-additive probability measures $\mu:\mathcal{P}(G)\rightarrow [0,1]$, so that $\beta G\subseteq \mathrm{Pr}^0(G)$. (Note that we consider every subset of $G$ to be measurable.) Given two probability measures $\mu,\nu\in \mathrm{Pr}^0(G)$, we should be able to convolve them as follows:
Construct the product probability measure $\mu\times \nu$, determined uniquely by the requirement that $(\mu\times\nu)(A\times B)=\mu(A)\nu(B)$ for all $A,B\subseteq G$.
Let the convolution $\mu*\nu$ be the pushforward of $\mu\times \nu$ via the multiplication map $G\times G\rightarrow G$. Thus $$(\mu*\nu)(E) := (\mu\times \nu)\left(\{(x,y)\in G\times G: xy\in E\}\right).$$
Is this correct? This should be an associative binary operation on $\mathrm{Pr}^0(G)$, making it into a semigroup.
Viewing $\beta G$ as a subset of $\mathrm{Pr}^0(G)$ by $p\mapsto \chi_p$, we can ask: is ultrafilter convolution the same as convolution of the corresponding probability measures?
Question: Does $p\mapsto \chi_p$ define a semigroup homomorphism $\beta G \rightarrow \mathrm{Pr}^0(G)$?