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I have just learned that Choi-Elliott proved in (ams.org/mathscinet-getitem?mr=1081290) that the set of self-adjoint elements with finite spectrum in certain rotation algebras are dense in the set of all self-adjoint elements. (Interestingly, this result is not explicit, in the sense that there is no concrete $\theta$ for which it happens.) However, from what I understand, it seems unclear whether there is a finitely generated dense sub-algebra such that each element has finite spectrum.
Another way of putting it is that every countably dimensional algebra carries a canonical locally convex topology which is called the fine topology. It is a topological algebra with this topology. Moreover, every ${\mathbb C}$-linear map into any other locally convex vector space is continuous. For finitely generated algebra, the pointwise convergence on the generators gives another way of defining the topology on the $Hom$-space.
What if you consider the group measure space construction of the action (of some element in) Thompsons group $F$ on $[0,1]$? There, you have an explicit element of $F$ which maps $[0,1]$ to itself (piecewise linear) such that $[0,1/2],[1/2,3/4],[3/4,1]$ is mapped to $[0,1/4],[1/4,1/2],[1/2,1]$. I think that this algebra (at least the crossed product by this specific automorphism) is a factor of type $III_{1/2}$. This should satisfy the pentagon axiom. In order to make everything (also the tensor product) work at the same time, more work is necessary.
I think that there is still some misunderstanding. The $\sum$-operation applied to two automorphisms would not be an automorphisms. (This is analogous to the fact that the sum of two line-bundles is not a line-bundle.) However, as far as I understand, your construction on the level of algebras would always yield that the sum of two automorphisms is an automorphism.
