If you want more of a "break free" effect, maybe have the force proportional to $1/r^3$ (inverse cube, like the force between two magnets) instead of the usual inverse square (gravity, electrostatic force). Or even $1/r^4$ or whatever you want. It can still scale linearly with mass the way gravity (${G M m}/{r^2}$) does, you just vary the scaling with distance.
That would also make it plausible for small nearby things (like fireballs) to have more of an effect than the Sun's pull. (Which only affects fire Elementals, not the Earth's orbit around the Sun, so would tend to pull fire Elementals off the surface during the day, counterbalanced by the pull of magma and stuff underground).
It's completely plausible for a force to have very different scaling with distance than gravity. Just don't call it gravity! For example, the Weak nuclear force is very short-range because the force-carrying particles have non-zero rest mass (unlike the photons that carry the electromagnetic force. This is why the electroweak force is unified at high energies: there is enough free energy for pair-production of the force carriers, so they can pop up like photons). I'm not saying you should go into that much detail (better to be vague than obviously wrong to people that know the subject), just that non-gravity-like forces happen without magic.
So you could even have a force that's more or less inverse-square over short ranges, but falls off dramatically at a certain distance.
There have also been proposed theories of gravity that modify it slightly over astronomical distances to explain things like the Pioneer (space probe) anomaly. (That's now pretty conclusively explained by thermal radiation pressure, ruling out some of the gravitational theories.)
Another plausible explanation for scaling different than inverse-square: extra spatial dimensions. i.e. direction(s) that Elemental stuff can move in that's perpendicular to x, y, and z. Inverse-square scaling happens because the area of a sphere scales with the square of radius. In 4-dimensional space, the analog of a sphere has a 3-D analog of area (actually a volume) that scales with the cube of radius.
So if the force (and force-carrying virtual particles) spread out in a 4th spatial dimension, it would scale as $1/r^3$. And you can still use the word "gravity" if you like.
Real theories have been proposed with properties like this, especially String Theory's extra dimensions which are "rolled up" / curved back on themselves, which limits the strength of the effect. One proposed effect is/was gravity being slightly weaker than $1/r^2$ over astronomical scales.
For this case, presumably it's a dimension that only Elementals can move in, not ordinary matter. So it can have a strong effect on Elementals without affecting normal physics. Perhaps moving in this dimension takes you between worlds, or between elemental planes and Earth.
Of course, it's easy to introduce inconsistencies if you aren't careful. (e.g. if there are lots of Elementals only a couple km away, they would exert some pull on Earth's air / water / fire / earth. (Assuming this force affects the inert element with an equal and opposite force to what the Elemental feels, otherwise you're violating conservation of momentum and conservation of energy.) Anyway, safer to just have the force carrying particles / fields spread out into the extra dimension(s), and not have the Elementals able to move along it. Otherwise they could bypass walls by going around them using this extra dimension, and stuff like that.
Remember that the shell theorem only holds with $1/r^2$ forces, so a short-range force will tend to pull Air elementals into the sky when they're on the ground. If it's stronger than gravity, they'd have to actively fly downwards or hold onto things to stay at surface level.
This Elemental force (E-gravity? gravitee? Etraction?) doesn't attract plain water to other plain water, for example. This means that buoyancy doesn't happen with respect to this force, because the displaced plain water wasn't affected by that force. (i.e. there's no extra pressure created by it).
We know this because Newtonian mechanics + fluid dynamics can accurately model the Earth's oceans (e.g. sea level heights around the world, and the deformation of the oceans due to Earth's spin rate). I think our models are detailed enough that we would have noticed if water attracted other water more than gravitationally, rather than just got the wrong value for some other free parameter.
Buoyancy due to gravity can overcome gravity + the extra force, if they're in the same direction, otherwise not.
Water density in the air is peanuts compared to a nearby ocean, even during a heavy rainstorm. A $1/r^3$ or $1/r^4$ force would make the effect of a storm more significant.
A water elemental leaving the ocean is like a human climbing a mountain. It's "uphill" all the way from the centre of mass of the ocean. Being at the edge of the ocean is like a human having climbed up a vertical shaft from the centre of the Earth.
However, the ocean is more like a thin disc than a sphere of water, so getting to the edge takes you farther from more of it. Unlike climbing outward from a sphere (where your weight will peak at the surface), your weight will probably drop off some as you approach the shore (especially since coastal water is shallower). However, with a $1/r^2$ force, the difference between getting to the edge and going another kilometre beyond is less significant than with a sphere. You're already pretty far from most of the water, so the total pull you feel doesn't drop off as much with distance from the shore. (So again, a $1/r^2$ force isn't going to give you much of a "break free of the ocean" effect).
If a Water elemental can "climb" more easily through water than by walking on land, it's going to be much easier to leave the ocean by going up-river until they're near an equilibrium between a large lake and the ocean.
