No. Define a function $f\colon S^1\to\{0,1\}$ by $f(x)=1$ if $x\in [0,\alpha)$ and 0 otherwise. Now define $\pi\colon S^1\to S^1$ by $\pi(x)=\sum_{n=0}^\infty f(R_\alpha^n x)m^{-(n+1)}$. It is easy to see that $\pi(R_\alpha x)=E_m(\pi(x))$. Now the measure defined on rectangles by $\mu(A\times B)=\lambda(\pi^{-1}(A)\cap B)$ is invariant under the product transformations. It is very far from a product measure: it is supported on the graph of $\pi$.

On the other hand... if you impose the condition that the product measure has Lebesgue measure the first marginal (that is $\mu(A\times S^1)=\lambda(A)$ for all measurable subsets of the circle), then the answer is yes. The property is called disjointness. Two measure preserving transformations are called disjoint if the only invariant measure for the product transformation that projects to each of the two measures under the coordinate projection is the product measure. This property was introduced in a famous paper of Furstenberg of 1967, Disjointness in Ergodic Theory.

Rudolph's book has a bunch of results about this, including Pinsker's result that any K process is disjoint from any process of entropy 0.