With some effort I can evaluate the integral in closed form:
$$I(a)=\int_{1}^\infty (x-\lfloor x\rfloor) x^{-a-1}\,dx=\frac{(1-a)\zeta (a)+a}{(a-1) a},\;\;a>0.$$
Hence the desired equality $\frac{f(a)-I(a)}{g(a)-I(1-a)}=1$ reduces to a simple equation for $f(a)-g(a)$:
$$f(a)-g(a)=\frac{\zeta (a)-a \zeta (1-a)-a\zeta (a)+2 a-1}{(a-1) a},\;\;0<a<1.$$