If you define inertia groups via valuation theory, then the exact same definition that works for finite extensions works also for infinite extensions. Namely, If $K/k$ is a (possibly infinite) Galois extension of global fields, and $v$ is place of $K$, then the inertia group, say $I(K/k)$, of $K/k$ consists of those $\sigma\in \operatorname{Gal}(K/k)$ such that
$$
v(\sigma(\alpha)-\alpha)>0
$$
for all $\alpha\in K$ satisfying $v(\alpha)\geq 0$. Incidentally, it is easy to check that this agrees with the other natural definition, that it is the inverse limit of the inertia subgroups $I(L/k)$ as you range over all finite galois subextensions $k/L/K$.
We are looking for a collection of ramification and splitting behaviours over a single prime $P$, all of which are possible, but rarely combine in a single easily-calculated example. I have produced a case where each of the things you ask for does indeed occur, in order to illustrate the general case. But I should point out that it is easy to produce individual lower-degree examples demonstrating separately each of the requirements you have placed on the primes above $P$.
Recall the well-known formula describing the splitting of a prime $P$ in an extension $L/K$ in terms of its ramification and residue field extension indices:
\begin{equation*}
\Sigma_{Q\mid P}\ e_{Q}f_{Q} = [L:K].
\end{equation*}
As you have pointed out, when the extension is Galois all of the $e_Q$ are equal to some fixed $e$, and all of the $f_Q$ are equal to a fixed value $f$; so this reduces to
\begin{equation}
efg = [L:K].
\end{equation}
where we have used the standard notation $g$ for the number of primes of $L$ above the prime $P$ of $K$ in a Galois extension.
We would like to see different types of behaviour at each of the levels $L$, $L_D$ and $L_E$ for some prime $Q$ of $L$ above $P$, which implies that we need $e\geq2$ and $f\geq2$. Moreover in order to have any chance of seeing different outcomes above the same prime $P$, we need that $P$ split into at least 2 distinct primes at the level $L_D$, otherwise $L_D=K$. Hence we also need $g\geq2$. So the degree of the extension $L/K$ needs to be at least 8.
Finally, as you also point out, we need that the Galois group Gal($L/K$) contain non-normal subgroups in order that any prime have a chance that the fixed field of its decomposition and/or inertia groups be non-Galois; otherwise we are just in a lower-degree version of the above formula. It is not sufficient, by the way, that the extension simply be non-Abelian, since for example the quaternion group Q8 is non-Abelian but has no non-trivial non-normal subgroups.
All of this forces the extension to have degree at least 16 (the non-normal subgroups of the dihedral group D4 of order 8 do not allow enough varied behaviour). We also would like an extension whose Galois group is furnished with lots of non-normal subgroups. So the easiest place to start would be something with Galois group S4. In order to simplify things let us assume that $K=\mathbb{Q}$, and take some "general" quartic extension, the simplest interesting one of which might be the splitting field $L$ of the polynomial $q(x) = x^4+x+1$.
I used MAGMA for the following calculations. This extension $L/\mathbb{Q}$ is ramified only above $P=229$, splitting into six primes with residue field extension (="inertia") degree $f=2$ and ramification degree $e=2$. Choosing any of these primes $Q$ say we have an inertia group $E(Q\mid P)$ which is cyclic of order 2 ($\cong C_2$) and a decomposition group $D(Q\mid P)$ which is isomorphic to $C_2^2$.
We then calculate that there is always a prime $Q_D'$ of $L_D$ over $P$ with $e_{Q_D'}=f_{Q_D'}=2$.
Similarly there is always a prime $Q_E'$ of $L_E$ over $P$ with $e_{Q_E'}=2$ and $f_{Q_E'}=1$.
Best Answer
I assume $L$ and $K$ are supposed to be number fields (or global fields). There are in general several decomposition (inertia) groups in $\mathrm{Gal}(L/K)$ attached to a finite prime $v$ of $K$, one for each finite prime of $L$ lying over $v$. If $w$ is such a prime of $L$, then the decomposition group $G_w$ is the subgroup of $\sigma\in\mathrm{Gal}(L/K)$ such that $\sigma w=w$ (if you think of $w$ as a prime ideal in the ring of integers in $L$, this means that $\sigma$ fixes the prime ideal, though not necessarily pointwise). Every element of $G_w$ gives rise to a well-defined automorphism of the residue field $k(w)$ of $w$, which can be thought of as $\mathscr{O}_L$ modulo the corresponding prime ideal. This gives a homomorphism from $G_w$ to $\mathrm{Gal}(k(w)/k(v))$. Its kernel is, by definition, the inertia group $I_w$ of $w$ in $\mathrm{Gal}(L/K)$. So the inertia group is a normal subgroup of the decomposition group. If $w^\prime$ is another prime of $L$ lying over $v$, then there is an element $\sigma\in\mathrm{Gal}(L/K)$ with $\sigma w=w^\prime$, and then the corresponding decomposition groups are conjugate (via $\sigma$). So, unless, e.g., there is only one prime of $L$ lying over $v$, the decomposition group $G_w$ will not be normal in $\mathrm{Gal}(L/K)$. The same goes for inertia groups.
To say that $v$ is unramified in $L/K$ in this context is to say that $I_w$ is trivial for every $w\mid v$ (or equivalently, for any $w\mid v$). So in this case the inertia group of $w$ is (trivially) a normal subgroup of $\mathrm{Gal}(L/K)$. When $G$ is abelian, conjugation by any element is the identity, so there really is a well-defined decomposition (resp. inertia) group attached to any prime $v$ of $K$, and they are of course normal in $\mathrm{Gal}(L/K)$.
I'm not really sure what you're asking about the absolute Galois group of $L$. This is a different object. Namely it is $\mathrm{Gal}(L_s/L)$ where $L_s$ is a choice of separable closure of $L$. The Galois group $\mathrm{Gal}(L/K)$ is can be regarded as a quotient of $\mathrm{Gal}(K_s/K)$ once an embedding of $L$ into $K_s$ is chosen. One can also define decomposition groups and inertia groups for infinite Galois extensions like $K_s/K$, and they have similar properties to the ones defined for finite Galois extensions. Also, if $\bar{w}$ is a prime of $K_s$ lying over the prime $w$ of $L$ which itself lies over $v$, then the image of the decomposition group $D(\bar{w})\subseteq\mathrm{Gal}(K_s/K)$ in $\mathrm{Gal}(L/K)$ is the decomposition group $D(w)$. The same goes for inertia groups.