pretty much.

yes, we are assuming e is an element of D. the fraction e/e (or d/d) acts as the unity of D^{-1}R (remember we are dealing with equivalence classes of RxD, not actual pairs (r,d). so there may be (and usually IS) MANY ways to represent "the same fraction". this is not much more than what we have in ordinary rational numbers:

3/4 = 6/8 = 9/12.....etc.)

this ring is usually written D^{-1}R, instead of R/D, to avoid thinking we are dealing of a "quotient ring" (that is, D is not necessarily an ideal of R).

the troubling inclusion of the "e" in re/de is only because 1/d may make "no sense", R may not, in fact, posses an element 1. if it does, we can use e = 1.

here is another example, which may give some additional perspective: suppose R = 2Z[x], all integer polynomials with even coefficients.

suppose we need to create a ring that contains this ring as a subring, but in which 2x^{4}has an inverse.

we can take D = 2Z[x^{4}] - {0}, non-zero polynomials in x^{4}with even integer coefficients (not closed under addition, but *is* closed under multiplication).

what is our "newly-created" inverse of 2x^{4}in D^{-1}R? well one possible form of it is:

2/(4x^{4}) (the top is in R, and the bottom is in D).

of course, for any p(x) in D, 2p(x)/(4x^{2}p(x)) would work just as well, for example, we could use:

p(x) = 2x^{4}+ 4, and in which case:

(4x^{4}+ 8)/(8x^{8}+ 16x^{4}) is also a form of (2x^{4})^{-1}.