If Poisson Process A, having intensity function lambda_A(t), is independent of Poisson Process B, having intensity function lambda_B(t), then A + B (which is not a mixture, but is what you intend from your stated failure rate context) is a Poisson Process having intensity function lambda_A+B(t) = lambda_A(t) + lambda_B(t).
See Fact 1.15 of http://www.actuarialseminars.com/Misc/PPjwd.pdf , and see derivation in homogeneous case at https://proofwiki.org/wiki/Sum_of_Independent_Poisson_Random_Variables_is_Poisson . I leave it as an easy exercise for you to show the result is true in the non-homogeneous case.
If the two Poisson Processes are not independent, there are many possibilities as to the distribution of their sum, and the calculations get more complicated. Independence is a critical assumption in failure analysis which can dramatically impact the result. Many people assume independence even when not justified, and as a consequence, their estimate of system failure rate (or probability) can be off by orders of magnitude. Maybe in your case independence actually holds, but you should carefully check (unless this is just a homework problem for which no real thinking should be applied).