An addendum to my comment to Qiaochu's answer, which amounts to expanding on Qiaochu's answer more generally: If we just add splitting of idempotents to our requirements on $(\mathcal{C}, \otimes)$, then $\mathrm{Comm}(\mathcal{C})$ is co-extensive! That is, the following equivalent conditions on a symmetric monoidal category $(\mathcal{C},\otimes)$ guarantee that $\mathrm{Comm}(\mathcal{C})$ is co-extensive:

 - $\mathcal{C}$ has finite biproducts and split idempotents, and $\otimes$ preserves finite biproducts in each variable separately.

 - $\mathcal{C}$ is enriched in commutative monoids and is Cauchy-complete and symmetric monoidal category in the enriched sense.

I find these conditions to be surprisingly mild and conceptual, compared to the definition of an extensive category, which I find a bit "fussy". I think these are pretty natural and minimal requirements of a category $\mathcal{C}$ to think of it as "a category of commutative modules", and being of the form $\mathrm{Comm}(\mathcal{C})$ for such a $\mathcal{C}$ is a pretty natural requirement for a category to be considered "a category of commutative algebras" -- although admittedly this is probably not the most natural way to think about commutative $C^\ast$-algebras. So the upshot is that any category of commutative algebras is dual to an extensive category, which is a good start toward being "a category of spaces", whatever that means.

But I don't know about the reverse direction -- given a "category of spaces", how likely is it to be dual to a category of algebras? This seems tricky, because there are certainly categories of spaces, such as projective varieties, which don't seem to be dual to categories of algebras -- we need to formulate some kind of "affineness criterion" in order to have a shot.

Anyway, it is not hard to show that $\mathrm{Comm}(\mathcal{C})$ is co-extensive when $\mathcal{C}$ is as above (it hardly can be, given how "clean" the hypotheses are!):

 - The 0-ary case of co-extensivity says that we have a strict terminal object (it admits no maps out except isomorphisms), which is true because the terminal object in $\mathrm{Comm}(\mathcal{C})$ is $0$, and if $0 \to A$ is a ring homomorphism, then the unit of $A$ is $0$ and the identity on $A$ factors through $0$.
 - The binary case says that the pushouts of the legs of a product decomposition again form a product decomposition. The product in $\mathrm{Comm}(\mathcal{C})$ is the biproduct in $\mathcal{C}$, so a map $A \times B \to C$ is induced by maps $f: A \to C$ and $g: B \to C$ in $\mathcal{C}$. Then composing $f$ and $g$ with the units on $A$ and $B$ we get commuting idempotents on $C$ which span $C$, and splitting the idempotents, we get a biproduct decomposition of $C$ in $\mathcal{C}$, which is a product decomposition in $\mathrm{Comm}(\mathcal{C})$.