Let $r$ and $s$ be positive real numbers, and let $A=(0,0)$, $B=(1,0)$, $C=(r,s)$, and $D=(r+1,s)$ be points on the coordinate plane. Find a point $P$, whose coordinates are expressed in terms of $r$ and $s$, with the following property: if $E$ is any point on the interior of line segment $AB$, and $F$ is the unique point other than $E$ that lies on the circumcircles of triangles $BCE$ and $ADE$, then $P$ lies on line $\overleftrightarrow{EF}$.
1. Let \( E = (k, 0) \). We will begin by calculating the coordinates of the circumcenters \( O_1 \) and \( O_2 \) of triangles \( ADE \) and \( BCE \), respectively.

2. First, consider the triangle \( ADE \):
   - The line segment \( \overline{AD} \) has slope \(\frac{s}{r+1}\) and midpoint \(\left(\frac{r+1}{2}, \frac{s}{2}\right)\).
   - The perpendicular bisector of \( \overline{AD} \) has slope \(-\frac{r+1}{s}\) and passes through \(\left(\frac{r+1}{2}, \frac{s}{2}\right)\). Its equation is:
     \[
     \frac{y - \frac{s}{2}}{x - \frac{r+1}{2}} = -\frac{r+1}{s} \implies y = -\frac{r+1}{s}x + \frac{r^2 + 2r + 1}{2s} + \frac{s}{2}
     \]
   - The perpendicular bisector of \( \overline{AE} \) is simply \( x = \frac{k}{2} \).
   - Substituting \( x = \frac{k}{2} \) into the equation of the perpendicular bisector of \( \overline{AD} \), we get:
     \[
     y = -\frac{r+1}{s} \cdot \frac{k}{2} + \frac{r^2 + 2r + 1}{2s} + \frac{s}{2}
     \]
   - Therefore, the coordinates of \( O_1 \) are:
     \[
     O_1 = \left( \frac{k}{2}, -\frac{r+1}{2s}k + \frac{r^2 + 2r + 1}{2s} + \frac{s}{2} \right)
     \]

3. Next, consider the triangle \( BCE \):
   - The line segment \( \overline{BC} \) has slope \(\frac{s}{r-1}\) and midpoint \(\left(\frac{r+1}{2}, \frac{s}{2}\right)\).
   - The perpendicular bisector of \( \overline{BC} \) has slope \(-\frac{r-1}{s}\) and passes through \(\left(\frac{r+1}{2}, \frac{s}{2}\right)\). Its equation is:
     \[
     \frac{y - \frac{s}{2}}{x - \frac{r+1}{2}} = -\frac{r-1}{s} \implies y = -\frac{r-1}{s}x + \frac{r^2 - 1}{2s} + \frac{s}{2}
     \]
   - The perpendicular bisector of \( \overline{BE} \) is \( x = \frac{k+1}{2} \).
   - Substituting \( x = \frac{k+1}{2} \) into the equation of the perpendicular bisector of \( \overline{BC} \), we get:
     \[
     y = -\frac{r-1}{s} \cdot \frac{k+1}{2} + \frac{r^2 - 1}{2s} + \frac{s}{2}
     \]
   - Therefore, the coordinates of \( O_2 \) are:
     \[
     O_2 = \left( \frac{k+1}{2}, -\frac{r-1}{2s}(k+1) + \frac{r^2 - 1}{2s} + \frac{s}{2} \right)
     \]

4. We wish for \( P = (x, y) \) to lie on the radical axis of the circumcircles of \( \triangle ADE \) and \( \triangle BCE \). By the Radical Axis Theorem, this is equivalent to \( P \) having equal power with respect to the two circles. Accordingly, we write:
   \[
   \text{Pow}_{(ADE)}(P) = O_1P^2 - O_1E^2 = \left( \frac{k}{2} - x \right)^2 + \left( y(O_1) - y \right)^2 - \left( \frac{k}{2} - k \right)^2 - \left( y(O_1) - 0 \right)^2
   \]
   \[
   \text{Pow}_{(BCE)}(P) = O_2P^2 - O_2E^2 = \left( \frac{k+1}{2} - x \right)^2 + \left( y(O_2) - y \right)^2 - \left( \frac{k+1}{2} - k \right)^2 - \left( y(O_2) - 0 \right)^2
   \]

5. Equating the powers and simplifying, we get:
   \[
   x^2 - kx + y^2 - \left( -\frac{r+1}{s}k + \frac{r^2 + 2r + 1}{s} + s \right)y = x^2 - (k+1)x + y^2 - \left( -\frac{r-1}{s}(k+1) + \frac{r^2 - 1}{s} + s \right)y + k
   \]

6. Further rearranging and simplifying, we obtain:
   \[
   x = \left( -\frac{r+1}{s}k + \frac{r-1}{s}(k+1) + \frac{r^2 + 2r + 1}{s} - \frac{r^2 - 1}{s} + s - s \right)y = \left( \frac{-2k + 3r + 1}{s} \right)y + k
   \]

7. Rewriting once more, we have:
   \[
   x - \frac{3r+1}{s}y = \left( -\frac{2y}{s} + 1 \right)k
   \]

8. In order for this equation to hold for all \( k \in [0, 1] \), it is necessary to have:
   \[
   x - \frac{3r+1}{s}y = -\frac{2y}{s} + 1 = 0
   \]

9. Solving these equations, we find:
   \[
   x = \frac{3r+1}{2}, \quad y = \frac{s}{2}
   \]

Thus, the coordinates of point \( P \) are:
\[
\boxed{\left( \frac{3r+1}{2}, \frac{s}{2} \right)}
\]