Proving that a language defined by a regular expression is equivalent to a right linear grammar
Let's try to simplify your grammar. First, notice that $A$ generates $(00)^*11S$, and so we can get rid of $A$, obtaining
$$
S \to 1(00)^*11S \mid 01S \mid \lambda
$$
Similar reasoning shows that your grammar generates the language
$$
(1(00)^*11+01)^* = ((1(00)^*1+0)1)^*,
$$
which is identical to your regular expression.
While the reasoning above is informal, it can be made formal with some work.