This answer is a little unsatisfying, as it amounts to: The two geometric constructions are equivalent, because the relations describing them are algebraically equivalent. I believe it's possible to remove some of the algebra, but I haven't found the cleanest way to do it ... yet!
We'll start with the two-circles construction:
Let circles $\alpha$ and $\beta$ have common center $O$ and respective radii $a$ and $b$. Let a variable ray from $O$ meet these circles at $A$ and $B$, and let the projections of $A$ and $B$ onto "horizontal" and "vertical" diameter-lines be $X$ and $Y$, and let the projection lines meet at $P$.
Defining $x := |OX|$ and $y:=|OY|$, the similarity of $\triangle OAX$ and $\triangle BOY$, and the right-triangle-ness of $\triangle OAX$, imply
$$\frac{|AX|}{|OA|} = \frac{|OY|}{|OB|} \quad\to\quad a^2 y^2 = b^2\;|AX|^2 \quad\to\quad a^2 y^2 = b^2 \left( a^2 - x^2 \right) \tag{1}$$
Of course, $(1)$ is equivalent to the "standard form" of the ellipse equation:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \tag{2}$$
Since we "know" $(2)$ derives from the focus construction, we could say we're done. But I want to investigate $(1)$ a little more.
Introduce $c$ such that $a^2 = b^2 + c^2$, and define $e := c/a$. Then $b^2 = a^2(1-e^2)$, and we can write $(1)$ as
$$\begin{align}
y^2 &= \left(1-e^2\right)\left(a^2-x^2\right)\tag{3} \\[6pt]
&= (a+ex)^2-(ea+x)^2 \tag{4}
\end{align}$$
Defining $z := ex$, we can write $(4)$ as
$$(a+z)^2 = y^2 + ( c+x )^2 \tag{5}$$
We interpret $(5)$ geometrically by introducing a third circle, $\gamma$, centered at $O$ and having radius $c$. Let $\gamma$ meet the variable ray at $C$ and the "horizontal" diameter-line at $C_{+}$ and $C_{-}$. Also, let $Z$ be the projection of $C$ onto that diameter. (While we're at it, let's say that the diamter meets $\alpha$ at $A_{+}$ and $A_{-}$.)
Now, $a=|OA_{+}|$ and $c=|OC_{+}|$, while proportionality tells us that $z = |OZ|$. We also have
$$y = |PX|,\qquad a+z = |A_{+}Z|,\qquad c+x = |C_{+}X| \tag{6}$$
so that $(5)$ implies, by way of right triangle $\triangle PXC_{+}$,
$$|A_{+}Z|^2 = |PX|^2 + |C_{+}X|^2 = |PC_{+}|^2 \quad\to\quad |A_{+}Z|=|PC_{+}| \quad\left(\text{likewise,}\; |A_{-}Z| = |PC_{-}|\right) \tag{7}$$
The reader may notice that @PhilH's expression for what he calls "$a$" amounts to the same observation. Even so, it's interesting (to me) to formulate things thusly:
Circle $\gamma$ meets the variable ray at a point whose projection onto the "horizontal" diameter of $\alpha$ divides that diameter into precisely the segments needed to join the foci to $P$.
I have no doubt that this interpretation exists in the literature; however, I only recently realized it, myself. Be that as it may ... We have recaptured the fact that the sum of the focus-to-$P$ distances is constant, namely the diameter of $\alpha$. $\square$
As I mentioned, I believe some of the algebra can be removed and replaced with more geometry. For instance, if we read equation $(5)$ as
$$y^2 = \left(\;(a+z)+(c+x)\;\right)\cdot\left(\;(a+z)-(c+x)\;\right) \tag{5a}$$
then we see that $y$ is the geometric mean of lengths $a+z+c+x$ and $a+z-c-x$; moreover, the arithmetic mean (aka, the average) of those lengths is $a+z$. These values and relations are geometrically (ahem) mean-ingful. (The "geometric mean" link shows how they feature in a classic construction involving a right angle inscribed in a semi-circle.) I have a way to deduce the relations (mostly) geometrically from the two/three-circle construction, but it's currently a little messier than the already-messy algebraic route. If (when?) I find a tidier argument, I'll update this answer.
