The papers of Zimmer are good references, but they only cover the commutative case (the case of an action on a measure space). For actions of (locally compact) groups on general von Neumann algebras, amenability is defined and studied by Claire Anantharaman-Delaroche in the following papers:
Claire Anantharaman-Delaroche,
Action moyennable d'un groupe localement compact sur une algèbre de von Neumann.(French. English summary)Math. Scand.45(1979), no.2, 289–304. https://www.mscand.dk/article/view/11844
Claire Anantharaman-Delaroche, Action moyennable d'un groupe localement compact sur une algèbre de von Neumann. II.(French. English summary)[Amenable action of a locally compact group on a von Neumann algebra. II] Math. Scand.50(1982), no.2, 251–268.
https://www.mscand.dk/article/view/11958
In the first paper it is already proved that the crossed product $M\bar\rtimes_\alpha\Gamma$ if injective if and only if $M$ is injective and the action $\alpha$ is amenable. Here $\Gamma$ is a discrete group. For general locally compact groups only one direction of this holds.
For actions of (discrete) groups on $C^*$-algebras, amenability is defined and studied in the follow up paper:
Claire Anantharaman-Delaroche, Systèmes dynamiques non commutatifs et moyennabilité.(French)[Noncommutative dynamical systems and amenability] Math. Ann.279(1987), no.2, 297–315. https://link.springer.com/article/10.1007/BF01461725
In this paper it is proved that for an action $\alpha$ of a (discrete) group $\Gamma$ on a $C^*$-algebra $A$, the crossed product $A\rtimes_\alpha \Gamma$ is nuclear if and only if $A$ is nuclear and $\alpha$ is amenable.
Amenable actions of locally compact groups were only defined and studied recently in the preprint (still not published):
https://arxiv.org/abs/2003.03469
Amenability and weak containment for actions of locally compact groups on $C^*$-algebras, by Alcides Buss, Siegfried Echterhoff, Rufus Willett
Further references, and historical background, can be found in that preprint.
For $x\in G,$ $x\neq e,$ the conjugacy class of $x$ is the subset of $G$ defined by
$$C_x=\{gxg^{-1}\:\, g\in G\}$$
The conjugacy classes do not change under the action of inner automorpisms. i.e. $hC_xh^{-1}=C_x$ for every $h\in G.$
According to Dietrich Burde comment
The von Neumann algebra $VN(G)$ of the group $G$ is a factor (the center of $VN(G)$ is trivial) if and only if for every $x\neq e$ the set $C_x$ is infinite.
For the proof of $\Rightarrow$ direction assume by contradiction that there is $x_0\in G,$ $x_0\neq e,$ such the set $C_{x_0}$ is finite. Consider the operator
$$A=\sum_{x\in C_{x_0}}\lambda_x$$ Clearly $A$ belongs to the algebra generated by $\lambda_g,$ $g\in G,$ as well as to its strong operator closure, i.e. to $VN(G).$ The operator $A$ commutes with translations $\lambda_h$ for every $h\in G.$ Indeed
$$\lambda_h A(\lambda_h)^{-1}=\lambda_h A\lambda_{h^{-1}}= \sum_{x\in C_{x_0}}\lambda_{hxh^{-1}}\underset{y=hxh^{-1}} {=}\sum_{y\in C_{x_0}}\lambda_y=A$$
The operator $A$ is nontrivial as $$\langle A\delta_e,\delta_{x_0}\rangle_{\ell^2(G)}=1$$ This completes the proof of $\Rightarrow$ direction.
For $\Leftarrow $ direction, assume that $VN(G)$ contains an operator $A,$ which commutes with all operators in $VN(G),$ hence it commutes with left translations $\lambda_g$ for every $g\in G.$ We are going to show that $A=0.$ Let $A\delta_e=\sum_{x\in G}a(x)\delta_x.$ The operator $A,$ restricted to the functions with finite support, is of the form
$$A=\sum_{x\in G}a(x)\lambda_x$$ as it commutes with right translations $\rho_g.$ Indeed
$$A\delta_y=A\rho_y(\delta_e)=\rho_y A\delta_e=\rho_y\left (\sum_{x\in G}a(x)\delta_x\right )=\sum_{x\in G} a(x)\delta_{xy}=\left (\sum_{x\in G} a(x)\lambda_x\right )\delta_y$$
As $A$ commutes with left translations we get
$$\sum_{x\in G}a(x)\lambda_x=A=\lambda_{g^{-1}}A\lambda_g=\sum_{x\in G}a(x)\lambda_{g^{-1}xg}=\sum_{x\in G} a(gxg^{-1})\lambda_x$$
Therefore the function $a(x)$ is constant on each conjugacy class.
Since $a(x)=A\delta_e\in \ell^2(G),$ the series
$\sum_{x\in G} |a(x)|^2$ is convergent. As each conjugacy class is infinite then $a(x)\equiv 0,$ i.e $A=0.$
Best Answer
Not entirely sure what you are looking for, since what $\lambda(1)$ is appears already in the text you quoted. You have $$ \lambda(1)\delta_n=\delta_{n+1}. $$ That is, $\lambda(1)$ is the bilateral shift corresponding to the canonical basis.