Competition between substitution and elimination (uni- or bimolecular)
For an $\mathrm{E1}$ mechanism, we would need to generate a carbocation. Our halide is attached to a secondary carbon so the generated carbocation would be secondary. A quick check of neighbouring stabilising groups is negative. And as a professor of mine once put it (possible slightly oversimplified but with some general truth in it):


  Exactly one secondary carbocation exists and that is the prop-2-yl cation.


Thus, $\mathrm{E1}$ is not possible since the carbocation will not form in any measurable amounts. For the same reason, $\mathrm{S_N1}$ is not an option. For everything else, let’s consider the molecule’s 3D structure:



The tert-butyl group acts as a conformation ‘anchor’ meaning that the other chair form is not viable. Luckily, bromine is pointing upwards, which gives us access both to $\mathrm{S_N2}$ and $\mathrm{E2}$ mechanisms without having to invert or use a boat conformation.

Our base/nucleophile is methanolate which is both a good enough base and a good enough nucleophile, able to access both pathways. The centre for nucleophilic attack is also well accessable. If I had to choose, I would indeed go with your choice of saying the main products will be $\mathrm{S_N2}$ adduct and two enantiomers of elimination product will also be formed. (They stem from either abstracting the proton I drew or the enantiotopic proton on the back half of the six-membered ring.)