…the set of points $P$ that is the answer is dependent on the distance between $A$ and $B$, thus I am to observe some property of the relationship between the sum of circles, which hardly makes sense to me…
Two points with this. Firstly, yes, the solution set depends on the distance $d_T{\left(A,B\right)}$, but it will scale uniformly, so you can set this to any distance you like without loss of generality.
Secondly, $d_T{\left(P,A\right)} + d_T{\left(P,B\right)}$ isn’t a “sum of circles”; it’s a sum of distances. The circle is the complete set expression.
This isn’t unique to taxicab geometry. A unit circle in standard Euclidean geometry could be stated as $\left\{P \mid d_E{\left(P,A\right)} = 1\right\}$, if we used $d_E$ for the Euclidean metric (Pythagoras’ theorem).
In fact, think about what $d_E{\left(P,A\right)} + d_E{\left(P,B\right)} = d_E{\left(A,B\right)}$—or in more ordinary notation, $\left\|P-A\right\| + \left\|P-B\right\| = \left\|A-B\right\|$—would mean: “The total distance between $P$ and both $A$ and $B$, is equal to the distance between $A$ and $B$.”
Since the points $A$ and $B$ are constant, so is the distance between them, and so we can say instead, “The total distance between $P$ and both $A$ and $B$ is a constant.” This is one definition of an ellipse. So we’re looking for the taxicab equivalent of an ellipse. A “circle” in taxicab geometry is a (Euclidean) square, so it makes a certain amount of sense that an ellipse would be a (Euclidean) rectangle—which is indeed the solution.
The fact that this specific choice of distance results in a degenerate ellipse—the line segment between $A$ and $B$—is a slight stumbling block, it’s true. But this happens because two (distinct) Euclidean circles can only intersect in one or two points, and requiring $P$ to be “as distant from $A$ and $B$ in total as they are from each other” results in the “one point” case.
In contrast, taxicab circles are Euclidean squares (with axis-aligned diagonals), and two (distinct) squares can intersect at one or two points, or along a partial or complete edge. If $A$ and $B$ lie on the same horizontal or vertical line, then once again we get the “one point” case. But if not…
…then the squares intersect along part of an edge, and these intersections form the perimeter and interior of a rectangle.