No.  The set of primes can be whatever you want (added: within reason!  As Benjamin Steinberg points out, it can in fact be any *recursive* set of primes).

First, note that for infinitely presented groups, the torsion can be whatever you like: the torsion in the group

$*_i \mathbb{Z}/p_i$

is precisely the set of primes $p_i$, by standard facts about free products.  (Added: As long as the set of $p_i$ is recursive, then this group admits a recursive presentation on a countable set of generators.)

By Higman's Embedding Theorem, the above group can be embedded in a finitely presented group.  More subtlely, this embedding doesn't introduce any new torsion---see, for instance, Theorem 2.5 of [this preprint of Chiodo][1].

**Clarification:** Higman's Embedding Theorem is commonly stated as only applying to finitely generated countable groups.  In fact, an old construction of Higman, Neumann and Neumann shows how to embed a countably generated group into a 2-generated group; if the countably generated group is recursively presented, then the 2-generated group can be taken to be recursively presented as well.


  [1]: http://arxiv.org/abs/1107.1489