The generic discrete logarithm problem is this:
Given a group $(G, ·)$ with generator $g$ and $y \in G$, find $x \in \mathbb N$ such that $y = g^x$.
The "classic" discrete logarithm problem (the one used in "classic" DSA and ECDSA) is this with some subgroup of the (multiplicative) group of a prime field, i.e. $(\mathbb Z/p \mathbb Z)^*$:
Given a prime number $p$, a generator $g$ (of some large subgroup) and $y \in \left<g\right>$, find $x \in \mathbb N$ such that $y = g^x \bmod p$.
The elliptic curve discrete logarithm problem is this with a subgroup of an elliptic curve group:
Given an elliptic curve $(E, +)$, a generator $G$ of some large subgroup, and $y \in \left<G\right>$, find $x \in \mathbb N$ such that $y = x · G$.
There are generic algorithms (such that the baby-step giant-step algorithm) for solving the discrete logarithm in any group (assuming a computable group law, or even just oracle access to the group law), and these work just fine for elliptic curve groups, as well as for the classic prime field. The average running time of these algorithms usually depends on the size of the group, like $O(\sqrt{|G|})$ for baby-step-giant-step. This is still exponential in the input size (since the input size is in $O(\log|G|)$, if one doesn't select really stupid encodings).
But for some special groups the discrete logarithm is easier, since they have some more (and known) structure. As an extreme case, consider the additive group $(\mathbb Z/n \mathbb Z, +)$ (for any positive integer $n$), with any generator. It is very easy to solve this problem even for quite large $n$ – we need only a single modular division, i.e. one execution of the extended euclidean algorithm. (And if the generator is fixed, we even can calculate its multiplicative inverse once and reuse it later, then it gets down to one multiplication per problem.)
The multiplicative group of the prime fields is harder. There is no known polynomial time algorithms, but there are still algorithms faster than the generic ones, using subexponential time. (Yes, there is some gap between exponential and polynomial.) The number field sieve for factoring integers can actually also be used to calculate discrete logarithms in prime groups (and needs similar running time for similar input sizes of both problems).
Thus we need larger prime field groups than we would need for a generic group for a same "security level".
For some elliptic curves, it looks like the discrete logarithm is almost as hard as for a generic group of the same size, which means that there are no specialized algorithms which are much faster than the generic ones. (When there are, one calls the corresponding group "broken". )
The effect is that for a similar security level, we can use quite smaller elliptic curve groups than prime groups (and it turns out that for these smaller groups, the calculation is also faster).
On the other hand, have a look at the RSA problem. It is about this:
Given $n$ (a product of two large unknown primes), $e$ (with $\operatorname{gcd}(m,e) = 1)$ and $x < n$, we look for $m$ such that $x = m^e \bmod n$.
The most efficient known method to solve this (other than maybe a brute force attack on $m$ when it comes from a small sub space) is by factoring the modulus, then calculating the private key $d$ and decrypt it. So the hard problem here is factoring the modulus.
It is not really clear how to generalize this to any group (or what other kind of structure). I found A semantically secure elliptic curve RSA scheme with small expansion factor (2002), and it refers to (and includes an description of) an earlier work (N. Demytko. A new elliptic curve based analogue of RSA. EUROCRYPT ’93,
LNCS 765 40–49 (1993).)
Demytko's scheme works on an elliptic curve (and its twists) over the ring $\mathbb Z/n\mathbb Z$ (where $n$ is still a product of two large primes), and encrypts using $c = e \star m$, which is the $x$-coordinate of $e·(m,y)$ for some $y$ such that this is on the curve, and decrypts a ciphertext $c$ as $m = d \star c$, where $d$ is one of four inverses of $e$, and it depends on $c$ which one to use.
The second one even works in a curve over $\mathbb Z/n^2 \mathbb Z$ (but claims to be semantically secure, as it incorporates a random component).
In both cases it looks like the easiest way to break this is to factor $n$ (in the natural numbers), not some analogous problem in an elliptic curve based structure.
This means $n$ must be just as large as for RSA, and thus we have correspondingly large curves – not really an advantage of using elliptic curves here.
