Centre of the algebra $\mathbb{Z} [\hat{W}]$ of the affine Weyl group For $W$ the Weyl group of some root datum $(P, R, P^\vee, R^\vee)$ (here $P$ is the weight lattice, $R$ the root lattice, and I will write the group structure additively) associated to an algebraic group $G$ and maximal torus $T$, the action of $W$ on $P$ gives the definition of the affine Weyl group to be $$ \hat{W} = W \ltimes P.$$
Now there is a natural isomorphism $\mathbb{Z}[P] \cong R(T)$ (the latter being the representation ring of the torus) given by sending the linear combination $\sum_{\lambda \in P} m_\lambda \lambda$ to the representation $V$ of $T$ which has a decomposition into weight spaces $V_\lambda$ with multiplicities $m_\lambda$.
This is unique up to isomorphism class.
Next, the canonical group embedding $P \hookrightarrow \hat{W}$ given by $\lambda \mapsto (1,\lambda)$ induces an embedding of group algebras $\mathbb{Z}[P] = R(T) \hookrightarrow \mathbb{Z}[\hat{W}]$, explicitly given by $$ \sum_{\lambda \in P} m_\lambda \lambda \mapsto \sum_{(1, \lambda) \in W \ltimes P} m_\lambda (1, \lambda).$$
Under this embedding, I can consider the space of $W$-invariants $R(T)^W \subset R(T)$, equivalently $\mathbb{Z}[P]^W \subset \mathbb{Z}[P]$ as the $W$-invariant linear combinations $\sum_{\lambda \in P} m_\lambda (1,\lambda) $ for which $$\sum_{\lambda \in P} m_\lambda (1, \lambda) = \sum_{\lambda \in P} m_\lambda (w, \lambda)$$ for any element $w \in W$ of the Weyl group.
This corresponds to representations $V$ of $T$ whose weight spaces are invariant under the action of $W$.
Here is my attempt at showing that

the centre of the algebra $\mathbb{Z}[\hat{W}]$ is $R(T)^W$ under the above embedding.

First, we know that the centre of the group ring $\mathbb{Z}[\hat{W}]$ is given by class functions on the affine Weyl group itself, that is, by linear combinations $ \sum f_{(w, \lambda)} (w,\lambda)$ for which $$\sum f_{(w, \lambda)} (w,\lambda) = \sum f_{(w, \lambda)} (s,\mu) (w,\lambda) (s^{-1}, -\mu)$$ for any $(s, \mu) \in \hat{W}$.
Unpacking the right-hand side using the definition of the group product in $\hat{W}$, this relation is equivalent to imposing that $$\sum f_{(w, \lambda)} (w,\lambda) = \sum f_{(w, \lambda)} (sws^{-1},\mu + s \cdot \lambda - sw \cdot \mu). \hspace{5em} (*)$$
First I want to show that elements of $R(T)^W$ satisfy the above relation $(*)$.
By $W$-invariance we have
$$
\sum m_\lambda (s1s^{-1},\mu + s \cdot \lambda - s \cdot \mu) = \sum m_\lambda (1,s^{-1}\mu + \lambda - \mu) 
$$
If I take $s = s_\alpha$ for $\alpha \in R$ a simple root (so that $s = s^{-1}$ is a simple reflection), this sum becomes $$ \sum m_\lambda (1,\lambda - \langle \mu, \check\alpha\rangle \alpha),$$
and it is here I am stuck - it isn't true that $\langle \mu, \check\alpha\rangle = 0$ for $\alpha$ in general, nor does it seem like I can move elements around in the same conjugacy class of $\hat{W}$.
I am equally stuck in the other direction: how can a class function of $\hat{W}$ look at all like $R(T)^W$?
 A: I prefer thinking of the extended affine Weyl group in terms of affine transformations, rather than an abstract semidirect product. For any $\lambda \in P$, let $T_\lambda \colon P \to P$ be translation by $\lambda$, so that $T_\lambda(\mu) = \lambda + \mu$. This embeds the additive group $P$ inside the group $\operatorname{Aff}(P)$ of invertible affine transformations of $P$. The Weyl group $W$ sits naturally inside $\operatorname{Aff}(P)$ as a set of linear transformations, then $\hat{W}$ is the group generated by $P$ and $W$.
This makes the semidirect structure natural: since $w \in W$ is a linear transformation it is easy to see that $w T_\lambda = T_{w \lambda} w$, leading to the multiplication rule
$$ (T_\lambda w)(T_\mu x) = T_{\lambda + w \mu} wx, \quad \text{or } (\lambda, w)(\mu, x) = (\lambda + w\mu, wx).$$
The inversion rule is also straightforward to deduce:
$$ (T_\lambda w)^{-1} = w^{-1} T_{-\lambda} = T_{- w^{-1} \lambda} w, \quad \text{or } (\lambda, w)^{-1} = (- w^{-1} \lambda, w^{-1}).$$
I bring this up because I think the formula for the inverse in the conjugation in the question is incorrect, since it only has $-\lambda$ rather than $-w^{-1} \lambda$.

First let's see that elements of $\mathbb{Z}[P]^W$ are central in the group ring $\mathbb{Z}[\hat{W}]$. Let $\sum_{\lambda \in P} a_\lambda e^\lambda \in \mathbb{Z}[P]^W$ be any element; we must have that $a_\lambda$ is finitely supported, and that $a_\lambda = a_{w \lambda}$ for all $w \in W$. Insert this into the group ring $\mathbb{Z}[P]^W$ to get $\sum_{\lambda} a_\lambda (T_\lambda)$. Now for a general element $T_\mu w$ we have
$$ T_\mu w \sum_\lambda a_\lambda (T_\lambda) = \sum_\lambda a_\lambda (T_{\mu + w\lambda} w) = \sum_\nu a_{w^{-1} \nu} (T_\nu T_\mu w) = \left(\sum_\nu a_\nu (T_\nu)\right) T_\mu w$$
where we used the substitution $\nu = w \lambda$ at the second equality, and the fact that $a_\lambda = a_{w \lambda}$ for the third. So these elements are indeed central.

Next, why is any central element of this form? Suppose $\sum_{\lambda, w} a(\lambda, w) (T_\lambda w)$ is central in $\mathbb{Z}[\hat{W}]$. To commute with a translation $T_\mu$, we must have
$$ \sum_{\lambda, w} a(\lambda, w) (T_\lambda w) = T_\mu \sum_{\lambda, w} a(\lambda, w) (T_\lambda w) T_{- \mu} = \sum_{\lambda, w} a(\lambda, w) (T_{\lambda + \mu - w \mu} w),$$
and hence $a(\lambda, w) = a(\lambda + \mu - w \mu, w)$ for all $\mu \in P$. If $w$ is not the identity then we can take some $\mu$ which is moved by $w$ so that $(1 - w)\mu \neq 0$. The same is true of $2 \mu$, $3 \mu$, and so on, so we get a chain of equalities
$$ a(\lambda, w) = a(\lambda + (1 - w)\mu, w) = a(\lambda + 2(1 - w)\mu, w) = \cdots$$
and since the coefficients $a(-, -)$ are finitely supported, sufficiently far out we find a zero, and so all these coefficients are zero. Therefore we have that $a(\lambda, w) \neq 0$ only for $w \neq \operatorname{id}$, so a central element must be of the form $\sum_\lambda a_\lambda T_\lambda$. You can easily check that requiring this sum to commute with the elements of $W$ forces the condition $a_\lambda = a_{w \lambda}$ for all $w \in W$, and so we are done.
