Asymptotic approximation of a recurrence relation (Akra-Bazzi doesn't seem to apply)
One possible approach might be by analogy to differential equations.  Let $T'(n) = T(n)-T(n-1)$.  Here $T'(n)$ is a discrete analog of the first derivative of $T(n)$.  We get the following relationship:
$$T'(n) = T(\lfloor \delta n \rfloor) + g(n).$$
The continuous analog of this is the differential equation
$$t'(x) = t(\delta x) + g(x),$$
or, if you prefer to see it written differently:
$${d \over dx} t(x) = t(\delta x) + g(x).$$
That's a differential equation.

Now you could try to solve the differential equation for the continuous function $t(x)$, then hypothesize that a similar function will be the solution to your original recurrence relation, and try to prove your hypothesis.  At least, this is one general approach you could follow.

I've forgotten everything I once knew about differential equations, so I don't know the solution that differential equation, but maybe you will be able to solve it by reviewing all the techniques for solving differential equations.