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If $p$ is odd and $E$ has good reduction at $p$, then the image of $K_p$ is independent of $E$ (i.e. does not depend on the particular $E$ other then requiring that it has good reduction). To be precise, the image will be $\mathbb Z_p^{\times}/(\mathbb Z_p^{\times})^2 Z_p^{\times})2 \times \mathbb Z_p^{\times}/(\mathbb Z_p^{\times})^2.$

Thus I don't think that there is not much chance that you will be able to extract any information about the global elliptic curve $E$ from knowing the image of $K_p$. (Even if $p = 2$ and/or the reduction is bad, there is very little information specific to $E$ in the image of $K_p$;

Have you looked at the discussion of descent and Selmer groups in (e.g.) Silverman's book? If you do, you'll see that the problem of doing descent involves looking at every prime (the point being that every prime intervenes in the definition of the Selmer group).

Concretely,in the case you are looking at, which I guess is an $E$ all of whose $2$-torsion is defined over $\mathbb Q$, what one sees is that if $P \in E(\mathbb Q)$, and we solve$Q = 2P$, then $Q$ is defined over a biquadratic extension of $\mathbb Q$ which is unramified away from $2$ and the primes of bad reduction. This greatly limits the possibilities, and is the basis for why descent works. In particular, the weak Mordell--Weil theorem --- i.e. the statement that $E(\mathbb Q)/2 E(\mathbb Q)$ is finite --- in this context is essentially the statement that there are only finitely many biquadratic extensions of $\mathbb Q$ with prescribed ramification.

If you focus on just one (or even a finite number) of $p$, you are throwing away this basic fact (i.e. that the field of definition of $Q$ is unramified away from finitely many primes).

2 typo

If $p$ is odd and $E$ has good reduction at $p$, then the image of $K_p$ is independent of $E$ (i.e. does not depend on the particular $E$ other then requiring that it has good reduction). To be precise, the image will be $\mathbb Z_p^{\times}/(\mathbb Z_p^{\times})2 Z_p^{\times})^2 \times \mathbb Z_p^{\times}/(\mathbb Z_p^{\times})^2.$
Thus there is not much chance that you will be able to extract any information about the global elliptic curve $E$ from knowing the image of $K_p$. (Even if $p = 2$ and/or the reduction is bad, there is very little information specific to $E$ in the image of $K_p$; it will just depend on generalities about the reduction type of $E$.)
If $p$ is odd and $E$ has good reduction at $p$, then the image of $K_p$ is independent of $E$ (i.e. does not depend on the particular $E$ other then requiring that it has good reduction). To be precise, the image will be $\mathbb Z_p^{\times}/(\mathbb Z_p^{\times})2 \times \mathbb Z_p^{\times}/(\mathbb Z_p^{\times})^2.$
Thus there is not much chance that you will be able to extract any information about the global elliptic curve $E$ from knowing the image of $K_p$. (Even if $p = 2$ and/or the reduction is bad, there is very little information specific to $E$ in the image of $K_p$; it will just depend on generalities about the reduction type of $E$.)