= Solution
All classes in the nontrivial homological equalities are taken in $H_1(X_{(p,q)};\mathbb Z)$, where $X_{(p,q)}=M_{S^2}(p^*/q,-q^*/p,*)$. The two internal gluing <torus> components still have inclusion maps into this <knot exterior>. Interpreting the maps instead as maps into the closed $S^3$ would make every displayed class zero.
Before the two <Dehn fillings>, $H_1(S^1\times\text{pair of pants};\mathbb Z)$ has generators $f,h_1,h_2,h_3$ and relation $h_1+h_2+h_3=0$. The <Dehn fillings> add $p^*f-qh_1=0$ and $-q^*f-ph_2=0$. Put $m=-h_3$. Eliminating the relations using $pp^*-qq^*=1$ gives
$$
\boxed{H_1(X_{(p,q)};\mathbb Z)=\mathbb Z\langle m\rangle,\quad
f=pq\,m,\quad h_1=pp^*m,\quad h_2=-qq^*m.}
$$
For completeness, the three relation rows in generators $(f,h_1,h_2)$ are $(p^*,-q,0)$, $(-q^*,0,-p)$, $(0,1,1)$ when computing the quotient by $m$. Their <determinant> is $pp^*-qq^*=1$, so $m$ really generates the entire <first homology group>, and no finite torsion or index is hidden in the elimination.
Orient each exceptional-fiber longitude by $\mu_i\cdot\lambda_i=1$. In the boundary basis $(m_i,\tilde f_i)=(-\tilde h_i,\tilde f_i)$, write $\mu_i=\alpha_i m_i+\beta_i\tilde f_i$ and $\lambda_i=u_i m_i+v_i\tilde f_i$, where $\alpha_i v_i-\beta_i u_i=1$. Then
$$
\tilde f_i=-u_i\mu_i+\alpha_i\lambda_i.
$$
The filled <meridian of a solid torus> maps to zero, and $(\alpha_1,\alpha_2)=(q,p)$, yielding
$$
\boxed{f=pq\,\iota_{3*}(m)=q\,\iota_{1*}(\lambda_1)=p\,\iota_{2*}(\lambda_2).}
$$
Adding multiples of $\mu_i$ to $\lambda_i$ has no effect. The word “any” therefore means any longitude with this compatible orientation; negating a longitude would negate the corresponding equality.
To compute <Turaev torsion>, use the general product formula $\tau(S^1\times F)\doteq(1-[f])^{-\chi(F)}$ and multiplicativity under gluing along <torus> components. For a <pair of pants>, $\chi(F)=-1$, so the unfilled piece contributes $1-[f]$. The two filling <solid torus> pieces contribute $(1-[\lambda_1])^{-1}$ and $(1-[\lambda_2])^{-1}$. With $t=[m]$, the equalities above give $[f]=t^{pq}$, $[\lambda_1]=t^p$, and $[\lambda_2]=t^q$. Therefore
$$
\boxed{\tau(X_{(p,q)})\doteq\frac{1-t^{pq}}{(1-t^p)(1-t^q)}.}
$$
Here $\doteq$ allows the unit $\pm t^n$: a fully refined <Turaev torsion> requires an <Euler structure> and a <homology orientation>, neither of which the statement specifies. The displayed rational function is the representative whose expansion at $t=0$ starts with $1$, equivalently the standard nonnegative-exponent normalization.
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