No - this is not true. The minimal example is provided by the deterministic Markov chain with two states (so that the state space is $\mathbb Z_2=\{0,1\}$) with the deterministic transitions $x\mapsto x+1\, (\!\!\!\! \mod 2)$. This chain is obviously ergodic in your sense with respect to the stationary measure $\mu$ with $\mu(0)=\mu(1)=1/2$. [I emphasize that your definition of ergodicity differs from the usual probabilistic definition, and rather coincides with what is called ergodicity in the theory of dynamical systems, see Different uses of the word "ergodic" .] Then the product Markov chain is also deterministic and has two orbits $\{(0,0),(1,1)\}$ and $\{(0,1),(1,0)\}$, so that it is not ergodic.

EDIT By the way, what is still true, is that if the tail $\sigma$-algebra of either chain is trivial (which is sometimes called ergodicity by probabilists), then the tail $\sigma$-algebra of the product is also trivial. The reason why this property does not imply the one you were asking about is precisely the same why the product of two ergodic (in the usual "dynamical" sense) transformations is not necessarily ergodic, and is related to **mixing**. Namely, in the Markov case the space of ergodic components of the chain is the quotient of its tail boundary (the one determined by the tail $\sigma$-algebra) by the time shift.