This is a good question, because it shows you're exploring the formulas yourself and questioning what you're told.
But I think you're attaching too much importance to a single expression in one particular form. If you look at derivations of the velocity-addition formula — at least the ones I know of — they just don't work when the velocity of the new frame is $c$ (either positive or negative). Instead, at some point they all involve division by $0$ or $\infty$ when the velocity of the new frame is $c$. It so happens that the form of the velocity-addition formula that you use has already been simplified to a state where there is no division by zero because the troublesome terms have already been eliminated. But by applying that form of the formula to this case, you've ignored the derivation and essentially applied an arbitrary formula to a physical situation where it doesn't apply. For example, what happens when this photon's "frame" tries to measure the speed of an object at rest in the original frame? You still get $-c$! (There's nothing special about the existence of this other photon.) Or how about a photon emitted in the same direction? Your formula gives $0/0$.
Ultimately, a frame is more than just an application of the velocity-addition formula to a single object; it is the entirety of the spacetime, measured with respect to a single state of motion and choice of origin. So, for example, to even define velocities you need some way to understand time and distance as measured in this photon's "frame". How do you do that? The standard way would be to involve the Lorentz transformations. But they are singular when $v=\pm c$ because of the factor of $\gamma = 1/\sqrt{1-v^2/c^2}$.
EDIT: There seems to be some confusion about the meaning of reference frames and transformations between reference frames, which I was assuming in the last paragraph above. A reference frame is simply a coordinate system, which gives a correspondence between some coordinate system and physical points of the spacetime. The velocity-addition formula doesn't actually involve coordinates directly, it just relates velocities — so it certainly can't "define" a frame. The suggested "frame" of the photon is not a valid coordinate system, so it cannot define a frame. I'll explain this in more detail, but in short, the problem is that you could convert your coordinates into the photon "frame", but you can never convert back from it — which means that that conversion is not a transformation.
The reason has to do with how frames transform. For simplicity, let's just start off with Minkowski spacetime and the usual Cartesian system of coordinates: $(t, x, y, z)$. One key feature of this coordinate system is that every single physical point in the spacetime is represented by some set of coordinates, and that set of coordinates only represents that one physical point. You can't find two separate physical points that are described by the same set of coordinates. [You might want to use some other coordinate system like spherical coordinates. In that case, you have a coordinate singularity at the North pole, for example, but that's not what I'm talking about. That's one physical point that's described by multiple coordinate points; I'm talking about multiple physical points described by the same coordinate point.]
Now, let's look at how these coordinates transform under a boost. Again, for simplicity let's just say the boost is in the $+x$ direction with some velocity $v<c$. The Lorentz transformation tells us how the coordinates transform into this new frame:
\begin{align}
t' &= \frac{1}{\sqrt{1-v^2/c^2}} \left(t - \frac{v\, x}{c^2} \right),
\\
x' &= \frac{1}{\sqrt{1-v^2/c^2}} \left(x - v\, t \right),
\\
y' &= y,
\\
z' &= z.
\end{align}
Now, these expressions involve division by zero if you try to take $v=c$, but you can take the limit of the expressions as $v \to c$, and you find that $t' \to 0$ and $x' \to 0$, regardless of the values of $t$ and $x$. But there's a crucial point about this limiting process: No matter how close $v$ is to $c$, different physical points with different values of $t$ and $x$ will have different values of $t'$ and $x'$, and you can always undo the transformation.
But that's not true if you extend the transformations to $v=c$ by just setting the new coordinate values equal to their limits. Instead, you've "flattened" the spacetime into two dimensions (neither of which is time, which makes it hard to understand what a velocity might be). In particular, all those physical points with the same $y$ and $z$ values but different $t$ and/or $x$ values are now described by the same coordinate point. So in this sense, it just doesn't make any sense to have a "frame" moving at velocity $v=c$ relative to another frame.
In principle, you could just go ahead and define this monstrosity, but good luck getting any physics out of it. For example, if you use this transformation on any vector, you'll get zeros in the $t$ and $x$ components. In any case, you're out of the realm of Special Relativity, and into the realm of your own theory.
