without more context, i can only put forth a guess, but i would surmise that a/b is meant to be an element of the field of fractions of D (for the integral domain Z, this is the rational numbers). we can form this field from any integral domain, in the same way we form rational numbers from integers.
in such a field, b(a/b) = a, so if a/b = c is in D, then a = bc in D (this is not, strictly speaking, correct. but there is a subring of the field of fractions isomorphic to D, and this isomorphism is often treated as equality. for example, we regard the fraction 4/2 as being an integer).
i daresay you could rephrase the theorem to read:
for a,b in D*, <a> = <b> iff a = bu, for some u in U(D).
i think that in this context, you could also take a/b in D to mean: "a is divisible by b". by analogy with the rational numbers: a is divisible by b iff a/b is an integer.
as i am not familiar with this text, all of the above is an educated guess (but it does fit the situation).
there is another use of a/b, with a similar construction based on a multiplicative subset S (all non-zero) of a commutative ring R, usually written S^-1R,
(sometimes R/S), of which the field of fractions of an integral domain D is a special case (clearly, D* is multiplicatively closed in an integral domain,
by definition). in this more general setting, R may not embed into S^-1R (if s is a zero-divisor in S, so that as = 0 for some a in R, then a/1 = as/s = 0. not good.).
if R is an integral domain, we happily avoid this unfortunate situation.