(Warning, ascii math is confusing and ambiguous to read. Sorr.)
Exponentiation of group "multiplication" does not immediately seem amenable to the reals, sure. But real exponentation does form a group, as shown here:
Define x_g(r) = the function that raises a Real/{0} (non-zero real) number r to the exponent x (in the sense of of some reasonable definition of exponentiaton of continuous functions). Define X = the set x_g() functions corresponding to all reals (including 0)
Define x_g y_g as composition: y_g(x_g(r)) = (r^x)^y = r^ (xy).
Then we have 0_g x_g = (r^0)^y = r ^ (0 y) = 1 = r ^ (y * 0) = (r^y)^0) = y_g 0_g -> identity
The group you described does not inherit any interesting structure from exponentiation -- indeed, one can easily see that it is isomorphic to the multiplicative group of reals. You could similarly construct a group isomorphic to an additive group of reals. This is an example of the fact that real exponentiation connects different aspects of real numbers, as well as the fact that just abstract algebra language is not enough to express properties of real numbers. You need to somehow relate the algebraic structure of reals to a topologic one, which stems from order imposed on reals and its continuity.
(Warning, ascii math is confusing and ambiguous to read. Sorr.)
Exponentiation of group "multiplication" does not immediately seem amenable to the reals, sure. But real exponentation does form a group, as shown here:
Define x_g(r) = the function that raises a Real/{0} (non-zero real) number r to the exponent x (in the sense of of some reasonable definition of exponentiaton of continuous functions). Define X = the set x_g() functions corresponding to all reals (including 0)
Define x_g y_g as composition: y_g(x_g(r)) = (r^x)^y = r^ (xy). Then we have 0_g x_g = (r^0)^y = r ^ (0 y) = 1 = r ^ (y * 0) = (r^y)^0) = y_g 0_g -> identity
y_g (1/y)_g = r^y ^ (1/y) = r^0 = 0_g -> inverse
(x_g y_g) (z_g) = ((r^x)^y)^z = (r^(xy))^z = r^((xy)z) = r^(x(yz) = (r^x)^(yz) = x_g (r_g r_g) -> Associativity
That makes a group.
Now, I explicitly left out the 0^x case. Can we fit it back in?
Not particularly cleanly, as thoroughly discussed in this thread.