Are there $2^{\aleph_0}$ pairwise non-isomorphic Boolean algebra structures on $\omega$?

The answer is yes: there are $2^{\aleph_0}$ countable Boolean algebras up to isomorphism, or equivalently $2^{\aleph_0}$ homeomorphism class of metrizable totally disconnected compact Hausdorff spaces. This is the main result of:

Reichbach, M. The power of topological types of some classes of 0-dimensional sets. Proc. Amer. Math. Soc. 13 1962 17-23 (Open link).

It precisely consists of showing that there are $c=2^{\aleph_0}$ closed subsets in a Cantor space, modulo global homeomorphism. Adding a discrete countable subset accumulating onto the given closed subset yields the desired family of continuum many non-homeomorphic metrizable Stone [=totally disconnected compact Hausdorff] spaces.

(Note that it also directly implies that there are $\ge c$ isomorphism types of Boolean subalgebras in $2^\omega$. At the topological level, classifying Boolean algebras embedding into $2^{\aleph_0}$ is the same as classifying [nonempty] separable Stone spaces. For Stone spaces, the class of metrizable spaces is properly contained in the class of separable ones. By Reichbach's 1962 result the former (modulo homeomorphism) has cardinal $c$ while by Freniche's 1984 result given by Juan, the latter has cardinal $2^c$.)

Another reference for "Answer = Yes":

Chapter 12 of the Handbook of Boolean algebras has the title "The number of Boolean Algebras".

Don Monk, the author of this chapter, writes: "For almost all classes K of BAs which have been an object of intensive study, there are exactly $2^\kappa$ isomorphism types of members of K of each infinite power $\kappa$."

In particular, this is true for K=interval algebras. There are $2^\kappa$ many linear orders $L$ of cardinality $\kappa$ such that the corresponding interval algebras $Int(L)$ are pairwise non-isomorphic. (The elements of $Int(L)$ are the finite unions of intervals of $L$.)

Here is an explicit sketch of Monk's proof (which he calls "of folklore nature"):

• Call a BA $B$ "atomic" if you can find an atom below every positive element, and call $x\in B$ "atomless" if there is no atom below $x$.
• For any BA $B$, let $I(B)$ be the (possibly improper) ideal generated by the atoms together with the atomless elements. (An ideal is improper if it is equal to the whole BA)
• Write $D(B)$ (the "derivative" of $B$) for the Boolean algebra $B/I(B)$.
• Let $A_0=\omega$ be the linear order of natural numbers, and let $A_1=1+\eta+\omega$ be the linear order of the nonnegative rationals followed by a copy of the natural numbers. Then $Int(A_0)$ is atomic, and $Int(A_1)$ is not, so you have two nonisomorphic algebras.
• (In both these linear orders we call the smalles element $0$.)
• Check that both $I(Int(A_0))$ and $I(Int(A_1))$ are improper, i.e. the derivatives of $Int(A_0)$, $Int(A_1)$ are singletons. This fact is responsible for the "crucial point" below.
• Now for any $i,j\in \{0,1\}$ consider $A_{i}\times A_{j}$ with the "inverse" lexicographic order (first compare two second components in $A_{j}$, and only when they are equal compare the first components)
• $A_0\times A_0$ and $A_0\times A_1$ lead to atomic interval algebras, but $A_1\times A_0$ and $A_1\times A_1$ do not.
• A crucial point is that the first factor "disappears" when taking the derivative: $D(Int(A_i\times A_j))$ is naturally isomorphic to $Int(A_j)$, and therefore atomic iff $j=1$. So the four interval algebras derived from $A_i\times A_j$ are pairwise nonisomorphic.
• This proof idea can be generalized to infinite (weak) products. For notational simplicity I restrict the proof sketch to countably many factors.
• For any sequence $x=(x(0),x(1),\dots)\in 2^\omega$ let $A_x$ be the weak product $\prod_i^{\rm wk} A_{x(i)}$, i.e., the set of all functions $f$ defined on $\omega$ which satisfy $f(i)\in A_{x(i)}$ for all $i$, and $f(i)=0$ for almost all $i$.
• The set $A_x$ is countable, and linearly ordered as an "inverse" lexicographic product.
• The main work to do is to check that the $n$-th derivative of $Int(A_x)$ is naturally isomorphic to $Int(\prod_{i\ge n}^{\rm wk} A_{x(i)})$, and hence is atomic iff $x(i)=0$.
• This show that for different $x,y\in 2^\omega$ the Boolean algebras $Int(A_x)$ and $Int(A_y)$ are not isomorphic.

Essentially the same proof can be used for uncountable $\kappa$, where the result is probably more interesting. (I seem to recall that all countable BAs are interval algebras.)

This was a problem in the Scottish Book posed by Ulam. It was solved by

F. J. Freniche, The Number of Nonisomorphic Boolean subalgebras of a Power Set, Proc. Amer. Math. Soc., 91 (1984) 199-201.

the number of non isomorphic sub algebras is $2^{2^k}$ for any infinite cardinal $k$