Do different methods of calculating fractional derivatives have to be equal?

The number of different kinds of fractional derivatives are many, from using laplace and fourier transforms, to integral transform definitions, to finite differences. Some are equivalent but some are different. See my answer here for a rundown. One of issues of fractional calculus, as I see it, is that there is no equivalent Bohr-Mollerup theorem that gives the most appropriate extension of differentiation like the theorem does for factorials.

But I want to make it clear that even in one particular definition, the Riemann-Liouville definition, we have to choose a reference point of integration in order to compute a fractional derivative or integral. In a bit I'll give you a concrete example.

The big jump from regular derivatives to fractional derivatives is that fractional derivatives are non-local operators. That is, $f'(c)$ is well defined and its value unchanged however $f$ is defined on $R \setminus B_{\delta}(c)$.

You are already aware of a non-local operator, which is itself a non positive integer derivative: integration, with $q=-1$. Consider the operator $[D^{-1} f](x) = \int_{-\infty}^x f(t) dt$. At $x=c$, any change in the value of $f$ in some positive measure set in $(-\infty, c)$ will potentially change the value of $[D^{-1} f](x)$. But if you look closely, we have a choice in the lower bound of $-\infty$. We could have just as well chosen $0$ or $-1$, or $1$.

Now you might think that this isn't a big deal because the difference is just a constant. Well, for integration that's true, but it gets more complicated with fractional differentiation.

The R-L definition of a fractional derivative is:

$$[D_a^{q}]f(x) = \frac{d}{dx^n}\frac{1}{\Gamma(n-q+1)} \int_a^x (x-t)^{1-q+n} f(t) dt $$ where $n=\max([q], 0)$ is the smallest nonnegative integer greater than $q$. (The reason we need to do these gymnastics is because the generalization of repeated integration has convergence issues for $q$ negative. So we integrate up fractionally and differentiate integerally to get a net differentiation of $q$.)

So now here is your example. Picking $a=-\infty$, one can compute using the R-L definition that

$$D_{-\infty}^q e^{bx} = b^q e^{bx},$$ for $b > 0$, while on the other hand $$D_{0}^q e^{bx} = x^{-q} E_{1, 1-q}(bx)$$

where $$E_{a,b}(x) = \sum_{k=0}^\infty \frac{x^k}{\Gamma(ak+b)} $$ is the Mittag-Leffler function. It would be a good exercise to compute these.

But actually, a simpler example to work out would be just to compute $$D_{a}^q 1 = \frac{(x-a)^{-q}}{\Gamma(1-q)},$$ and realize the dervitative cannot be equal for different choices of $a$ even up to a constant and more so, even up to a polynomial for non-integer $q$.

The question might now be this: Instead of making equivalence classes of functions modulo constants to make integration unique, can we make some new equivalence class of functions modulo some condition to make fractional differentiation unique? I am not sure this is possible, but it is worth investigating.