Allow me to elaborate on Benjamin's answer:
Proposition: There is a fixed finitely presented group $H$, and fixed finite set of words $S$ on the generators of $H$ for which $\langle S\rangle \leq H$ has solvable subgroup membership problem, such that the problem of determining if the subgroup generated by one word $\langle w \rangle$ intersects $\langle S \rangle$ non-trivially in $H$ is algorithmically undecidable.
Proof: There exists a finitely presented group $G=\langle X|R \rangle$ for which the word problem is solvable, but for which the torsion problem, of determining if a word has infinite order, is unsolvable. This is a consequence of Theorem A in D. Collins, The word, power, and order problem in finitely presented groups, in Word Problems, eds. W. W. Boone, F. B. Cannonito and R. C. Lyndon, 401–420 (1973). (On top of this theorem, one needs the additional observation that having solvable word problem means one can effectively compute the order of a torsion element). So now take the group $H$ to be $F_{X} \times F_{X}$, and take the set $S$ to be the finite set $\{(x_{1}, x_{1}), \ldots, (x_{n}, x_{n}), (r_{1}, 1) \ldots, (r_{m}, 1)\}$, and thus $\langle S \rangle =\{(u,v) \in F_{X}\times F_{X}\ |\ u=v \ \ in \ \ G \}$. Since $G$ has solvable word problem, we have that $\langle S \rangle \leq H$ has solvable membership problem. Moreover, as Benjamin observed, for an arbitrary word $z \in G$, we have $\langle (1,z) \rangle$ intersects $\langle S \rangle$ trivially if and only if $z$ has infinite order in $G$ (the latter being undecidable).
I suppose the above should really be a comment to Benjamin's answer, but at the time I couldn't add comments as I had <50 reputational points.
-Maurice