A matrix with entries in ED is invertible if and only if the matrix can be row reduced to the identity matrix 
A matrix with entries in an Euclidean Domain is invertible if and only if the matrix can be row reduced to the identity matrix.
Is the above statement true?

 A: By the Smith normal form for Euclidean domains, for any given matrix $ X $, there is $ A $ and $ B $; both products of matrices corresponding to invertible elementary row/column operations, such that $ X = ADB $ for a diagonal matrix $ D $. (Stated this way, the Smith normal form relies essentially on the Euclidean algorithm, so is not true for general principal ideal domains.) If $ X $ is invertible, it follows that the nonzero entries of $ D $ are all units, which we may wlog assume to be $ 1 $. Therefore; $ X = AB = (AB)I $, where $ AB $ is a product of matrices corresponding to invertible elementary row operations, and therefore $ (AB)^{-1} X = I $. Thus, the answer to the question is positive.
Things are not as trivial as Josué Tonelli-Cueto surmises in the comments; as using the Euclidean algorithm during row reduction may cause elements which have been reduced to zero to become nonzero again. The Smith normal form, however, gives a straightforward way of answering the question. (Note that the algorithm for producing the Smith normal form also allows for column reduction, so the problems we may run into are resolved more easily.)
