One particle states in an interacting theory

I'll address your issues with definition (1):

$E$ is a function of $\vec p$ because $\lvert \lambda_{\vec p}\rangle\sim\lvert \lambda_0\rangle$ where by $\sim$ I mean that they are related by a Lorentz boost. That is, to "construct" these states, you actually first sort out all the states $\lvert \lambda_0\rangle,\lambda_0\in \Lambda$ ($\Lambda$ now denotes the index set from which the $\lambda_0$ are drawn) with $\vec P \lvert \lambda_0\rangle = 0$ and $H\lvert \lambda_0\rangle = E_0(\lambda)\lvert\lambda_0\rangle$ and then you obtain the state $\lambda_{\vec p}$ by applying the Lorentz boost associated to $\vec v = \vec p/E_0(\lambda)$ to $\lvert\lambda_0\rangle$. Since $H$ and $\vec P$ area components of the same four-vector, that means that $E_{\vec p}(\lambda)$ is a function of $E_0(\lambda)$ and $\vec p$ - and $E_0(\lambda)$ is determined by $\lambda$, so $E_{\vec p}(\lambda)$ is really a function of $\vec p$ and $\lambda$.

Conversely, every state of momentum $\vec p$ can be boosted to a state with zero momentum. We examine this zero-momentum state for what $\lambda_0$ it is and then name the state we started with $\lambda_p$, so this really constructs all states.

From the above it also directly follows that $E_{\vec p}(\lambda)^2-\vec p^2 = E_0(\lambda)^2$. I'm not sure what your issue with degeneracy is - degeneracy in $P$ and $H$ is not forbidden - nowhere is imposed that $E_{\vec p}(\lambda) \neq E_{\vec p}(\lambda')$ should hold for $\lambda\neq\lambda'$.

In completely precise, nonperturbative terms, a one particle state in an interacting QFT is a state that belongs to an irreducible invariant subspace of the Hilbert space of the interacting representation of the Poincare group corresponding to an isolated shell in the spectrum of the translation subgroup.

A one particle momentum state $|p\rangle$ is an eigenstate of the four components of the momentum operator (translation generators in the representation of the Poincare group) that belongs to such a subspace.

One-particle states are the eigenstates of the mass-squared operator which are orthogonal to the vacuum.

Multi-particle states correspond to the continuous part of the mass spectrum .

Electrons however are infraparticles, see en.wikipedia.org/wiki/Infraparticle .

And for quarks it is not clear what mass means since they are confined.