You could let $g(x) = 2$ if $x\geq 0$, and $g(x)=-2$ if $x<0$. Then $\int\limits_{-a}^a \frac{f(x)}{g(x)}\,dx = \int\limits_{-a}^0 \frac{f(x)}{-2}\,dx + \int\limits_{0}^a \frac{f(x)}{2}\,dx = 2\int\limits_{0}^a \frac{f(x)}{2}\,dx = \int\limits_{0}^a f(x)\,dx.$
The condition actually isn't very hard to satisfy: all you need is for $|g(x)|$ to be bounded below and for $\frac{1}{g(x)}-\frac{1}{g(-x)}$ to equal $1$ identically; then the exact same reasoning goes through.
In the even case, the similar condition $\frac{1}{g(-x)}+\frac{1}{g(x)}=1$ is satisfied with $g(x)=1+e^x$, since $\frac{1}{1+e^{-x}} + \frac{1}{1+e^x} = \frac{(1+e^x) + (1 + e^{-x})}{(1+e^x)(1+e^{-x})} = \frac{1+e^x + 1 + e^{-x}}{1+e^x + e^{-x} + 1} = 1.$
If you want something similar for the odd case, there's no such nonzero $g(x)$ that's also continuous at $x=0$. That's because if $\frac{1}{g(x)}-\frac{1}{g(-x)}=1$ (which you can prove using bump functions to be necessary, and not just sufficient), then letting $x\to 0$ we have $0=\frac{1}{g(0)}-\frac{1}{g(0)} = 1.$ So you might as well use the discontinuous $g$ described above.