I am very far from an expert, so I hope my answer is useful for you, but I hope also that people who know more than I do will supplement or correct my answer.

Let me fix a commutative ring (or...) $k$. The first thing to recall is that the Brauer group of $k$ is very close to the etale homology group $\mathrm H^2(\operatorname{spec} k, \mathbb G_m)$. For example, when $k$ is a field, then etale homology is group homology for the Galois group, and (writing $k^s$ for the separable closure of $k$) one has $\text{Brauer} = \mathrm H^2(\operatorname{Gal}(k^s/k),(k^s)^\times)$, where $(k^s)^\times$ is the group of units in $k^s$. Grothendieck asked (and at least in some cases answered, but I don't know the history) whether one actually has Brauer=H^2 in general, where "Brauer" is defined by Azumaya algebras and corresponds geometrically to projective bundles over $\operatorname{spec} k$ (modulo proj(vector bundles)). The answer is that one does have an inclusion of Brauer into H^2, but it is not onto in general. First, every Azumaya algebra represents a torsion element in H^2, and in general H^2 has non-torsion part. But in fact there can be torsion in H^2 that also is not in Brauer. To fix these problems, see for example Heinloth and Schröer, "The bigger brauer group and twisted sheaves", 2009. (Sorry, I don't have an arXiv number, but I can email the pdf if you can't find it online.) Perhaps these are the infinite-dimensional algebras you're looking for. If memory serves, I learned this story from Pete L. Clark here on MathOverflow. (If memory has misserved me, I apologize!)

There is a story I have told myself, but I don't know a reference or let alone a precise statement. Let me be imprecise about some list of "nice categorical properties", and wrap everything up into the word "green". ("green" may include words like "presentable" or "abelian" or ....) Then for example the category $K = \text{Mod}_k$ of right $k$-modules is a green category. Because $k$ is commutative, $K$ has a natural structure as symmetric-monoidal category. For any algebra $A$ over $k$, one can also consider the green category $\text{Mod}_A$ of right $A$-modules. Any such module is also a left $k$-module, and so there is an operation of $k$-modules on $A$-modules, and this makes $\text{Mod}_A$ into some categorified version of *left $K$ module*. In general, we can consider a 2-category whose objects are green categories equipped with the structure of left $K$-module; I will call this 2-category $_K\text{MOD}$.

Then since $K$ is symmetric monoidal, $_K\text{MOD}$ carries a natural structure as symmetric monoidal 2-category. After you've decided what "green" means, a natural question to ask is: What are the "units" in $_K\text{MOD}$? For comparison, the group of units in $K = \text{Mod}_k$ are precisely the line bundles over $\operatorname{spec} k$, i.e. the Picard group $\mathrm H^1(\operatorname{spec} k,\mathbb G_m)$. So this is a sort of "2-Picard group", and the "0-Picard group" is $\mathbb G_m(k) = \mathrm{GL}_1(k) = \mathrm H^0(\operatorname{spec} k,\mathbb G_m)$.

When "green" means "is the representation theory of an algebra", probably this "2-Picard group" is the classically-defined Brauer group on the nose. What you would like to believe (but as I said, I have no theorem to this effect) is that for the correct definition of "green", the 2-Picard group is precisely $\mathrm H^2(\operatorname{spec} k,\mathbb G_m)$. Then the answer to your question is that rather than looking for infinite-dimensional algebras that extend the Brauer group, what you should be asking for are representation theories that behave as if they were representations of an "infinite-dimensional" central simple algebra. And you should believe that the story continues for the correct categorification of "green".

But then I think that even for the correct definition of "green", over a field you don't get anything new. That's because we already have Brauer = H^2 on the nose, and so you should believe that all reasonable definitions of "green" have the same group of units. Most generally, there is a weakening of "unit" to "dualizable object" and probably when $k$ is a field, all reasonable definitions of "green" give the same class of dualizable objects in $_K\text{Mod}$. But I'm starting to digress far from your original question, and so I will stop.