Here is my attempt to address Eric's actual question. Given a real $n$-dimensional vector bundle $E$ on a space $X$, there is an associated Thom space that can be understood as a twisted $n$-fold suspension $\Sigma^E X$. (If $E$ is trivial then it is a usual $n$-fold suspension $\Sigma^n X$.) In particular, if $E=L$ is a complex line bundle, it is a twisted double suspension. In particular, if $X = \mathbb{C}P^\infty$, the twisted double suspension of the tautological line bundle $L$ satisfies the equation $$\Sigma^L\mathbb{C}P^\infty = \mathbb{C}P^\infty.$$ As I understand it, Eric wants to know whether this periodicity can be interpreted as a Bott map, maybe after some modification, and then used to prove Bott periodicity. What I am saying matches Eric's steps 1 and 2. Step 3 is an attempted modification to make the map look a little more like Bott periodicity.
I think that the answer is no. It's an interesting map, but as I see it, it does not carry the same information as the Bott map. The homotopy information in the domain, the target, and the map all change a lot. Of course, with enough new ideas you could try to change the map to make it the Bott map, but I think that such an effort would be stone soup.
First, I think that the Bott map as Bott constructed it in the Annals is beautiful. The map generalizes the suspension relation $\Sigma S^n = S^{n+1}$. You do not need Morse theory to define it; Morse theory is used only to prove homotopy equivalence. Bott's definition: Suppose that $M$ is a compact symmetric space with two points $p$ and $q$ that are connected by many shortest geodesics in the same homotopy class. Then the set of these geodesics is another symmetric space $M'$, and there is an obvious map $\Sigma M' \to M$ that takes the suspension points to $p$ and $q$ and interpolates linearly. For example, if $p$ and $q$ are antipodal points of a round sphere $M = S^{n+1}$, the map is $\Sigma(S^n) \to S^{n+1}$. In the compelx cases, Bott uses $M = U(2n)$, $p = q = I_{2n}$, and geodesics equivalent to the geodesic $\gamma(t) = I_n \oplus \exp(i t) I_n$, with $0 \le t \le 2\pi$. The map is then $$\Sigma (U(2n)/U(n)^2) \to U(2n).$$ The argument of the left side approximates the classifying space $BU(n)$. Bott show that this map is a homotopy equivalence up to degree $2n$. In explicit terms, if a space $X$ has an $n$-plane bundle given by a map to the Grassmannian $\text{Gr}(n,2n)$, then there is also a Bott map $$\Sigma X \to U(2n).$$ Of course, you get the nicest result if you take $n \to \infty$. Also, to complete Bott periodicity, you need a clutching function map $\Sigma(U(n)) \to BU(n)$, which exists for any compact group.
(If you apply the general setup to $M = G$ for a simply connected, compact Lie group, Bott's structure theorem shows that $\pi_2(G)$ is trivial; c.f. this related MO question.)
Eric's twisted suspension only exists for $\mathbb{C}P^\infty = BU(1)$, not $BU(\infty)$. You can compute that $\pi_2(\mathbb{C}P^\infty) = \mathbb{Z}$ and the other homotopy groups are trivial. So you do not need Bott's theorem to understand the homotopy type of this space, and in any case there is no periodicity. If $X$ is a space with a line bundle $L$, then it has a map to $\mathbb{C}P^\infty$. Eric thus constructs a map $$\Sigma^L X \to \mathbb{C}P^\infty,$$ which means that the twisted suspension also has a line bundle $L$. You can check that $\Sigma^L X$ depends on the choice of $L$. For instance, if $X = S^2$ and $L$ is trivial, then $\Sigma^L S^2 = S^4$ is the usual suspension. But if $L$ has Chern number 1, then $\Sigma^L S^2 = \mathbb{C}P^2$, as Eric computed. Moreover, the purpose of the suspension maps in spectra and Bott periodicity is to go from a map $\Sigma X \to Y$ to its adjoint map $X \to \mathcal{L}Y$. In this case, since it is a twisted suspension, you instead get a bundle over $X$ with fiber $\mathcal{L}^2\mathbb{C}P^\infty$. Again, this could be interesting, but it is not Bott periodicity.
Finally Eric's step 3 formally desuspends both sides to get $$\Sigma^{-2}\Sigma^L \mathbb{C}P^\infty \to \Sigma^{-2}\mathbb{C}P^\infty.$$ You can't desuspend unless you first suspend infinitely many times to get a spectrum. However, the spectrum $\Sigma^{\infty}\mathbb{C}P^\infty$ takes you on a yet different track. This is a really complicated spectrum reminiscent of the sphere spectrum. If $\mathbb{C}P^\infty$ is too simple to resemble $BU(\infty)$, $\Sigma^{\infty}\mathbb{C}P^\infty$ is too complicated to resemble it.