All right, I'd approach this in the following manner.
Part 1: True Anomaly $\theta$ at $r$ in terms of Elliptical Orbit Eccentricity $e$
The first place I'd start is the Vis-Viva Equation, which, for all orbits around a particular body with a particular gravitational parameter $\mu$ links relative speed $v$ with radial distance $r$ and semimajor-axis $a$.
$$v^2 = \mu\left ( \frac{2}{r}-\frac{1}{a} \right )$$
From this we know that if $v_{ellipse}$ = $v_{circular}$ = $v$ at $r$, then $r$ = $a$ for both the ellipse and the circle.
Your provided equations for True Anomaly $\theta$ for elliptical orbits goes to the following once we substitute in $r$ for $a$. We'll just grab the one in the first or second quadrant.
$$\theta = \cos^{-1} \left ( \frac{r(1-e^2)}{re} - \frac{1}{e}\right )=\cos^{-1} \left ( \frac{(1-e^2)}{e} - \frac{1}{e}\right )$$
$$= \cos^{-1} \left ( -\frac{e^2}{e} \right ) = \cos^{-1} (-e)$$
Part 2: Flight Path Angle $\gamma$ at $r$ in terms of the Elliptical Orbit Eccentricity $e$
Flight path angle, $\gamma$ (at the link it's "Angle of velocity relative to the perpendicular to the radial direction):
$$\gamma = \tan^{-1}\frac{e \sin \theta}{1 + e \cos\theta}$$
And using trigonometric identities, I can get to (though I'm being a bit sloppy with my $\pm$):
$$\gamma = \tan^{-1}\frac{e \sqrt{1-\cos^2 \theta}}{1 + e \cos\theta} = \tan^{-1}\frac{e \sqrt{1-e^2}}{1 - e^2} = \tan^{-1}\frac{e }{\sqrt{1-e^2}}$$
And thanks to HopDavid for reminding me of inverse trigonometric identities, because that means for the flight path angle,
$$\gamma = \sin^{-1}e$$
Part 3: $\Delta v$ required to go from circular to elliptical orbit at $r$
And, as mentioned by David Hammen in the comments, you can't express the delta-v solely in terms of $e$ and $r$, because the delta-V is absolutely going to be a function of $v$, and that's going to be dependent on what you're orbiting, and how far you are from it.
If you have the Flight path angle, the speed, you can then call on the Law of Cosines to acquire the delta-V, as we're just looking at vector addition with two legs of magnitude $v$ separated by an angle $\gamma$
$$\Delta v^2 = 2v^2 - 2v^2 \cos\gamma = 2v^2 - 2v^2 \cos\left({\sin^{-1}e}\right) $$
$$
= 2v^2 \left(1 - \cos\left(\sin^{-1}e\right)\right)
$$
Inverse trigonometric identity to the rescue again, and:
$$
\Delta v^2 = 2v^2(1 - \sqrt{1-e^2})
$$
Thanks again to HopDavid for pointing out my missing square on the Law of Cosines. That brings us to
$$
\Delta v = \sqrt{2v^2(1 - \sqrt{1-e^2})}
$$
At the very least, if we are allowed to bring in the gravitational parameter $\mu$, we can go to the Circular Orbit Velocity equation:
$$v = \sqrt{\frac{\mu}{r}}$$
And wind up with the following.
$$\Delta v = \sqrt{2\frac{\mu}{r} \left(1 - \sqrt{1-e^2}\right)}$$
but I'm uncertain if that's allowed by the letter of the question. This agrees with HopDavid's answer and I thank him for his assistance.