Answer to the quick version. Yes it is true as soon as $(X,\mu)$ is a Lebesgue space. Beware that the transformation on the product $A_i\times B_i$ is not necessarily a true product, but instead it is a skew-product of the form $(x,y)\mapsto (T_x(y),y)$. This follows from the ergodic decomposition theorem, together with the classification of measurable partitions.
Recall that if T is an invertible measurable transformation acting on a Lebesgue space X, then there is a measurable partition $C_i$ (which may have uncountably many elements) and
probability measures $\mu_i$ on $C_i$ such that all $C_i$ are invariant by T, T is ergodic w.r.t $\mu_i$ and $\mu$ is obtained by integrating the $\mu_i$.
$$\mu(A) = \int_X \mu_i(A) d\mu$$
There are two kinds of ergodic components $C_i$. The one of positive measure, there are at most countably many such components. Let us remove these components from $X$, together with the periodic points, which are easily dealt with. Rohlin structure theorem on measurable partitions (1947) now says that
there is a isomorphism between $([0,1]\times [0,1], \lambda)$ and $(X,\mu)$ such that the pullback of the measurable partition $(C_i)$ is the decomposition into horizontal lines $([0,1]\times \{i\})_{i\in [0,1]}$. A reference is the book of Parry, "entropy generators in ergodic theory".
Here is how the ergodic decomposition is often used. If it happens that a result is true for an ergodic transform, then it is true for an arbitrary transform in restriction to its ergodic components, and you (may) recover the result on the whole space $X$ just by using the integral formula given above.
A reference for the ergodic decomposition for countable groups action is Glasner, "ergodic theory via joinings" th. 3.22.
Finally the result you are alluding in your last question is a section theorem. Given a measure preserving transform between two Lebesgue spaces X and Y, there is a section from Y to X, up to null sets, and some warning is in order here because this is not true in the Borel category. I think this is again due to Rohlin, and it can be deduced from its structure theorem for measurable partitions. Have a look at the book of Parry, but really this is overkill.
EDIT: following the comments of R.W., here is a counterexample to having a true product, instead of just a skew-product. On $[0,1]\times [0,1]$ take
$(x,y)\mapsto (x+y\ \ mod\ \ 1,y)$, together with Lebesgue measure. The restrictions to the fibers $[0,1]\times \{y\}$ are ergodic for a.e. y, and give uncountably many different isomorphic systems, as can be shown by looking at their spectra.
Your proof can be modified a bit so that it works for a general countable, amenable group $\Gamma$. In this case, we can take $B(X)$ to be the closure of $\mbox{span} \{ f \circ T_{\gamma} - f : f \in C(X), \gamma \in \Gamma \}$ and we show that $B(X)$ doesn't contain the constant functions on $X$ as follows:
I claim that every function $h \in \mbox{span} \{ f \circ T_{\gamma} - f : f \in C(X), \gamma \in \Gamma \}$ is non-negative at some point in $X$, namely $\max_{x \in X} h(x) \ge 0$. This is sometimes called the "translate property" for amenable groups, if I'm not mistaken. Anyway, after establishing that, the conclusion follows easily: it follows that one cannot uniformly approximate a negative constant (say $-7$) by a function of the above form.
The proof of the translate property I'm familiar with goes like this:
Assume for contradiction that $\max_{x \in X} h(x) = \delta < 0$ and let $\varepsilon > 0$. Let $h=\sum_{i=1}^{n}f_{i}\circ T_{\gamma_{i}}-f_{i}$. Since $\Gamma$ is amenable we can find a finite set $U \subset \Gamma$ such that $\frac{\left|\gamma_{i}U\triangle U\right|}{\left|U\right|}<\varepsilon$ for all $i=1,\dots,n$ (the Følner condition). Now consider the function $F:=\frac{1}{\left|U\right|}\sum_{u\in U}h\circ T_{u}$. Choose some point $x_0 \in X$. On the one hand, by our assumption, $F\left(x_{0}\right) = \frac{1}{\left|U\right|}\sum_{u\in U} h \left(T_{u}x_{0}\right)\le\delta<0$. On the other hand,
$\left|F\left(x_{0}\right)\right|=\left|\frac{1}{\left|U\right|}\sum_{u\in U}\sum_{i=1}^{n}f_{i}\left(T_{\gamma_{i}}T_{u}x_{0}\right)-f_{i}\left(T_{u}x_{0}\right)\right|$
By the triangle inequality, changing order of summation and using the property of U, you see that the above is at most $2n \varepsilon \max_{x\in X,1\le i\le n}\left|f_{i}\left(x\right)\right|$. So if you choose $\varepsilon$ small enough, you get a contradiction.
Best Answer
The answer is yes, such an action exists.
What is needed for the construction is the following very nice example of an action of a non-amenable group on $\mathbb Z$, which I just learned from Gabor Elek.
Consider a graph with vertices given by $\mathbb Z$ and unoriented edges between $n$ and $n+1$.
Pick a random labelling of the edges by the letters $a,b$ and $c$ with no $a$, $b$ or $c$ adjacent to the same letter. This defines an action of the group $G=\mathbb Z/2 \mathbb Z \ast \mathbb Z/2 \mathbb Z \ast \mathbb Z/2 \mathbb Z$. Indeed, just act according to existing labels or fix the element.
This action has the nice feature that it keeps invariant all counting measures on $\mathbb Z$, i.e. all $\mathbb Z$-Folner sequences sets are also Folner sequences for the $G$-action.
Now, the space of labellings (as above) of the graph is itself a probability measure space (a Bernoulli space), which carries an ergodic p.m.p. $\mathbb Z$-action by shifting. It is easy to see that $G$ acts on this space by measure preserving transformations (just by the method described above, done orbit by orbit) and induces an action as required. Indeed, the orbits are just the $\mathbb Z$-orbits, so its ergodic and amenable. Faithfulness follows the fact that you considered all labellings, so that with positive probability (on the space of labellings), an element will act non-trivially. Note also that $G$ is not amenable.
EDIT: As requested, more details on the action. The elements of the shift space are maps $f: \mathbb Z \to \lbrace a,b,c \rbrace$ with $f(n) \neq f(n+1)$. A letter shifts $f$ to the right if $f(1)$ equals that letter, it shifts to the left, if $f(0)$ is equal to the letter; otherwise you fix $f$. It is obvious that the orbits are just the orbits of the shift-action of $\mathbb Z$. Hence, the induced equivalence relation is just the one induced by the action of $\mathbb Z$.