Where does the energy of the destructively interfering light waves go in this interferometer setup? In one type of Mach-Zander interferometer two parallel light guides are diffused into a $\text{LiNbO}_3$ wafer. There are electrical contacts deposited along each light guide. At the end of the light guide both light guides are combined in to a single light guide. A voltage applied to the contacts can be used to change the light propagation speed in each light guide. 
Light is propagated through each light guide. Using voltages applied to the electrical contacts the light waves in the two light guides are adjusted in such a way that the light waves cancel where the light guides are combined. In this case no light emerges from the single light guide at the end of the interferometer.
We never observed a significant amount of light being back reflected from the interferometer.
What happened to the energy of the light beams?
 A: As it is shown in this sketch from Wikipedia

there are two out-coming images, the one is the negative of the second.
A experimental setup, showing the same phenomenon, this time with the back reflection from one mirror, see on YouTube from MIT OpenCourseWare. The details from the 5:30 min.
A: A diagram would confirm your setup, but I am assuming the recombination is into a symmetric Y-junction, where the two input guides and the output guide all have only one bound mode. So the short answer to your question is that the "lost" light is coupled into the radiation modes of the output waveguide in the destructive interference case, shed into the chip and is therefore not output through the waveguide itself. It can help to imagine the Y-junction to be like a single moded $2\times2$ coupler where one of the output arms simply goes into a beam dump.
Further insight is afforded by the Lorentz reciprocity law, which shows that any linear reciprocal $N$ port must have a symmetric scattering matrix. So imagine the $Y$ junction as a splitter where unit input power into the output waveguide were split into powers $p$ and $1-p$ with $0<p<1$ out of the two input guides. The relevant $S$-matrix elements are then $e^{i\,\phi_1}\,\sqrt{p}$ and $e^{i\,\phi_2}\,\sqrt{1-p}$ (there are phase factors $\phi_j$ in general). Given the $S$-matrix's symmetry, this means that if unit amplitude is input into each of the two input guides separately the amplitude of the output guide in each case would also be $e^{i\,\phi_1}\,\sqrt{p}$ and $e^{i\,\phi_2}\,\sqrt{1-p}$ in each separate case, respectively, with the lost power coupled into radiation modes. Again, it can help to imagine the splitter as a lossless $2\times2$ coupler with one arm connected to a beam dump to infer the correct behavior. The general output amplitude with phase $\phi$ between the arms is then:
$$\exp\left(i\,\left(\frac{\phi}{2}+\phi_1\right)\right)\,\sqrt{p}+\exp\left(i\,\left(-\frac{\phi}{2}+\phi_2\right)\right)\,\,\sqrt{1-p}$$
