A recursive definition of a function (or anything else for that matter), is given from one ore more "recursive equations", with the name of the recursive function ('f' in your example) as "the unknown". Now if that equation(s) has a unique solution, then we can say that the function is defined (uniquely determined) from said equation(s); we can now think of the equation(s) as some sort of "implicit definition".
It has nothing to do with recursion or functions per se; while $x = 2$ or $3x=0$ "define" numbers (2 and 0 respectively), the equations $0\cdot x=0$, $x=x$ and $0\cdot x=17$ do not (either because no solution exists as in $0\cdot x=17$ or the solutions are not unique).
Now the thing with recursive definitions is, that it can be shown, that when certain conditions are met (roughly "there is a well founded relation underlying the recurrence"), then a unique solution for the given equation exists.
In your example $f(x)=f(x+1)$, the underlying relation is the "greater than" relation, which is not well founded and indeed no unique solution for that equation exists (as stated all constant functions satisfy the equation) and thus it does not qualify as a "definition".