There is an old result of Tarski which says that any algebra A of sets which is
- κ-complete (i.e. A is closed under unions and intersections of fewer than κ sets from A), and
- satisfies the κ-chain condition (i.e. there is no family of κ many pairwise disjoint nonempty sets in A),
then A is necessarily atomic.
The proof of atomicity is along the lines of what Jonas suggested, but with a wellordered descending chain. Given a0 ∈ A and x ∈ a0. Starting from a0, recursively construct a strictly descending wellordered chain (aζ: ζ < δ) of sets in A, each containing x, for as long as possible. Since the differences aζ \ aζ+1 are pairwise disjoint and nonempty, this construction must terminate with δ < κ. Therefore, the intersection a = ∩ζ<δ aζ is an element of A by κ-completeness and x ∈ a. This intersection must in fact be an atom in A, otherwise we could extend the chain further. It follows that a0 is the union of all atoms below a0 and hence that A is atomic.
In your case, you have an ω1-complete algebra of sets. If the underlying set of the algebra is countable, then there certainly cannot be a family of ω1 many pairwise disjoint nonempty sets in A. Note also that the case κ = ω shows that finite algebras of sets are atomic.
It is possible to find the following: A compact abelian group G with Haar measure $\mu_G$, a subset $S\subseteq G$ of full outer Haar measure and a measurable function $f\colon S\to X$ with $\mu_P(E)=\mu_S(f^{-1}(E))$ for measurable $E\subseteq P$. In fact, as you mention, G can be taken to be a large enough product of the circle group.
Here, I am implicitly referring to the sigma algebra $\mathcal{B}(S)\equiv\{S\cap E\colon E\in\mathcal{B}(G)\}$ and $\mu_S$ is the restriction of the Haar measure to $\mathcal{B}(S)$, $\mu_S(S\cap E)=\mu_G(E)$. Ideally, we would like to enlarge S so that it is actually of full measure, then it could be enlarged to all of G. I'm not sure if this is possible though (and suspect that it is not possible in ZFC). The problem is that if $f\colon P\to Q$ is a measure preserving map of probability spaces then $f(P)$ will be of full outer measure, but need not be measurable. If $f^{-1}\colon\mathcal{B}_Q\to\mathcal{B}_P$ is an onto map of their sigma algebras then P and Q are "almost isomorphic" probability spaces. If, however, $f(P)$ is not measurable then f does not have a right inverse (even up to zero probability sets), unless we restrict to the subset $f(P)\subseteq Q$.
In the following, I write 2={0,1}, so that, for a set I, 2^I is the set of {0,1}-valued functions from I.
Letting $\pi_i\colon2^I\to\{0,1\}$ be the projection onto the i'th coordinate, the sets of the form $\pi_i^{-1}(S)$ generate a sigma algebra, which I will denote by $\mathcal{E}_I$.
First, we can reduce the problem to that of measures on $2^I$.
Step 1: Given a collection $\{A_i\}_{i\in I}$ of sets generating the sigma algebra on P, construct a probability measure $\mu_I$ on $2^I$ and a measure preserving map $f\colon P\to 2^I$ such that $f^{-1}\colon\mathcal{E}_I\to\mathcal{B}_P$ is onto.
Then, f will have a right inverse $g\colon f(P)\to P$ which is automatically measurable and measure preserving.
To construct f, define $f(p)\in 2^I$ by $f(p)(i)=1$ if $p\in A_i$ and =0 otherwise. It can be checked that $\mu_I(S)=\mu(f^{-1}(S))$ for $S\in\mathcal{E}_I$ satisfies the required properties.
Now, I will use $G^I=(\mathbb{R/Z})^I$, which is a compact abelian group with Haar measure $\mu_{G^I}$, which is the product of the uniform measure on the circle $G=\mathbb{R/Z}$.
Step 2: Construct a measure preserving map $f\colon G^I\to 2^I$.
Once this map is constructed, putting it together with step 1 gives what I claimed.
For any $J\subseteq I$, use $\pi_J\colon 2^I\to2^J$ and $\rho\colon G^I\to G^J$ to denote restriction to J. Also use $\mu_J(S)=\mu_I(\pi_J^{-1}(S))$ for the induced measure on $2^J$.
Zorn's lemma guarantees the existence of a subset $J\subset I$ and measure preserving map $f\colon G_J\to 2^J$ which is maximal in the following sense: for $J\subseteq K\subseteq I$ and measure preserving $g\colon G^K\to2^K$ with $\pi_J\circ g=f\circ\rho_J$ then K=J.
Then, J=I and we have constructed the required map. If not, choose any $k\in I\setminus J$, $K=J\cup\{k\}$ and define $h_k\colon G^K\to 2$ as follows (I let $A_k=\pi^{-1}_k(1)\subseteq2^I$).
$$
h_k(x)=\begin{cases}
1,&\textrm{if }0\le x_k\le\mu_I\(A_k\mid\mathcal{E}_J\)\_{f(x)}\\\\
0,&\textrm{otherwise}.
\end{cases}
$$
Defining $g\colon G^K\to2^K$ by $g(x)_i=f(x)_i$ for $i\in J$ and $g(x)_k=h_k(x)$ gives a measure preserving map extending f, and contradicting the maximality of J.
Best Answer
You write "An example of a σ-algebra that has no atoms but supports a probability measure is $\{0,1\}^\kappa$ for $\kappa$ uncountable, which we can endow with the coin-flipping probability measure."
Maharam's theorem says that these are essentially the only ones. That is: Every Boolean algebra which is equipped with a probability measure (and is Dedekind complete, see below) is isomorphic to a product of the measure algebras on various $2^\kappa$ that you mentioned. (Including finite $\kappa$, to take care of measures with atoms.)
Dedekind complete means that every subset has a least upper bound. If you take a $\sigma$-algebra which carries a $\sigma$-additive probability measure, and divide by the ideal of null sets, then the resulting algebra is still a measure algebra and it will be Dedekind complete.
An exposition of Maharam's theorem can be found in Fremlin's book, volume 3. (The theorem I quoted can be generalized to algebras with a "semifinite" measure, which is more general than probability measure.)