Solution:

The locus sought is the union of the open segment $P Q$ and the part of the circle with diameter $P Q$ that lies outside the strip bounded by $r$ and $s$.
Each circle can be tangent to its respective line in two ways: in one case it intersects the strip $S$ between the two lines, in the other it does not intersect it at all. In both cases the center of the circle (resp. $O_{P}, O_{Q}$) lies along the perpendicular to the line through the point of tangency. Moreover, if the circles are tangent at $T$, the line through the two centers passes through $T$.
For the pair of circles this leads us to 4 cases: we immediately exclude the one in which neither of the two circles intersects $S$, because in this case the circles cannot be tangent to each other.
Suppose first that both intersect $S$: then $T$ belongs in turn to $S$. The angles $P \hat{O}_{P} T$ and $T \hat{O}_{Q} Q$ are alternate interior angles with respect to the two parallel lines $P O_{P}$ and $Q O_{Q}$ (which are such because both are perpendicular to $r$ and $s$) cut by the transversal $O_{P} O_{Q}$, and are therefore equal. They are the central angles of the two isosceles triangles $P O_{P} T$ and $T O_{Q} Q$, which therefore also have equal base angles: from $O_{P} \hat{P} T=O_{Q} \hat{Q} T$ one finally deduces that $P, T$ and $Q$

are collinear, that is, that $T$ is a point interior to the segment $P Q$.
Conversely, if $T$ is a point interior to the segment $P Q$, let $O_{P}$ be the point of intersection of the perpendicular bisector of $P T$ with the perpendicular to $r$ passing through $P$, and let $O_{Q}$ be the point of intersection of the perpendicular bisector of $Q T$ with the perpendicular to $s$ passing through $Q$. By the similarity of the isosceles triangles $P O_{P} T$ and $Q O_{Q} T$, one has,
similarly to before, that $T$ lies on the segment $O_{P} O_{Q}$. By drawing the circle with center $O_{P}$ tangent to $r$ and the circle with center $O_{Q}$ tangent to $s$, one thus finds that $T$ lies on the line of centers and belongs to both circles, so that the two circles are externally tangent at $T$.
Suppose now that only one of the two intersects $S$: by symmetry, suppose that this happens for $C_{P}$. The point $T$ then belongs to the (open) half-plane $H$ bounded by $s$ not containing $r$. Let $R$ be the point symmetric to $Q$ with respect to $O_{Q}$. The tangent at $R$ to $O_{Q}$ is parallel to $s$ and hence to $r$; by what was shown in the first case, $P, T$ and $R$ are collinear. But since $Q$ and $R$ are diametrically opposite, $Q T$ is perpendicular to $P R$. But then $T$ must belong to a semicircle with diameter $P Q$, in addition to $H$.

Conversely, if $T$ is a point of the half-plane $H$ belonging to the semicircle with diameter $P Q$, we construct, as in the first case, $O_{P}$ as the point of intersection of the perpendicular bisector of $P T$ with the perpendicular to $r$ passing through $P$ and $O_{Q}$ as the point of intersection of the perpendicular bisector of $Q T$ with the perpendicular to $s$ passing through $Q$. The verification that the circle with center $O_{P}$ and radius $O_{P} P$ and the circle with center $O_{Q}$ and radius $O_{Q} Q$ are tangent at $T$ is the same as that of the first case.
The locus sought is therefore the union of the open segment $P Q$ and the part of the circle with diameter $P Q$ that lies outside the (closed) strip bounded by $r$ and $s$.

Observing as before that $P O_{P} T$ and $Q O_{Q} T$ are isosceles triangles and that their vertex angles $P \hat{O}_{P} T$ and $Q \hat{O}_{Q} T$ are supplementary, one obtains by difference that the sum of the base angles of these two triangles are complementary. It follows that $P \hat{T} Q=180^{\circ}-P \hat{T} O_{P}-Q \hat{T} O_{Q}=90^{\circ}$. From here one concludes as in the previous solution.