This is a fundamental group of a graph of groups with B-S groups as vertex groups and cyclic groups as edge groups. So yes, each Baumslag-Solitar group naturally embeds in your group.

** Edit.** It is not a graph of groups but a complex of groups: triangle with B-S groups at vertices, cyclic groups as edges and 1 in the triangle. So you need to check the Haefliger condition to prove that it is developable. One can also (better) use Stallings, John R.
Non-positively curved triangles of groups. Group theory from a geometrical viewpoint (Trieste, 1990), 491–503, World Sci. Publ., River Edge, NJ, 1991.

** Edit 2.** No, this method does not work. To compute the Gersten-Stalling angle, one needs to count only the alternating length, not the total length of the word. The alternating length of each relator is not 7 but 4, so each angle is $\pi/2$ and the total angle is $3\pi/2$, too large.

** Edit 3.** In fact the group is finite, so none of the B-S groups embeds. See this paper (the link was sent to me by Dani Wise).