Question: "I've been trying to prove it for quite a while now, but haven't made any progress. Does anybody know a proof of this or a resource to find one?"
Answer: Let $k$ be a fixed commutative unital ring and let $G:=Spec(R)$ be a $k$-group scheme
($R$ is a finitely generated $k$-algebra).
Let $\epsilon: R \rightarrow k$ be any $k$-valued point $\epsilon \in G(k)$ with $I:=ker(\epsilon)$ and let
$$\mu\in Hom_k(R/I^{n+1}, R/I):=Dist^n(R,I):=Dist^n(G,I).$$
It seems to me that there is an isomorphism
$$Dist^n(R,I) \cong Diff^n_k(R/I^{n+1}, R/I)$$
where $Diff^n_k(R/I^{n+1}, R/I)$ is the module of $n$'th order differential operators from $R/I^{n+1}$ to $R/I$.
You may see this as follows: There is a splitting $s:R/I\rightarrow R/I^2$ and let $a:=s \circ p$ where $p: R/I^2 \rightarrow R/I$ is the canonical map. Any element $u\in I^2$ is a sum of elements on the form
$$u:=(x-a(x))(y-a(y))z$$
with $x,y,z \in R$. Since $\mu$ is well defined we get
$$A1.\text{ }0=\mu((x-a(x))(y-a(y))z)=$$
$$\mu(xyz)-x\mu(yz)-y\mu(xz)+xy\mu(z)=[[\mu, \phi_x] \phi_y](z)=0$$
where $\phi_x(w):=xw$ is the multiplication map. Hence $\mu$ is a differential operator of order one. An induction proves the result for all $n\geq 1$.
Definition: If $k\rightarrow R$ is a map of commutative rings and $E,F$ are $R$-modules you define $Diff^n_k(E,F)$ inductively as the set of maps $D\in Hom_k(E,F)$
with
$$[\cdots [D, \phi_{a_0}]\cdots ]\phi_{a_{n}}]=0$$
for all elements $a_0,..,a_n\in R$. Here $[D,\phi_a](u):=D(au)-aD(u)$.
Hence the map $\phi_a$ is multiplication with $a$ and the product $[,]$ is the Lie product. By definition
$$Diff_k(E,F):= \cup_n Diff_k^n(E,F).$$
You must verify that your map $D_{\mu}$ satisfies the conditions in the definition.
You get a canonical map
$$ \phi:R\otimes_k R \rightarrow R/I^{n+1}\otimes_k R \rightarrow^{\mu \otimes 1}R/I\otimes_k R \cong R$$
defined by
$$ \phi(a\otimes b):= \mu(\overline{a})b \in R$$
which is $k$-linear. Let $J\subseteq R\otimes_k R$ be the ideal of the diagonal and let $J^{n+1}$ be its $n+1$'th power. It seems to me that $\phi(J^{n+1})=0$ in $R$ hence you get a canonical $k$-linear map
$$\phi_{\mu}: R\otimes_k R/J^{n+1} \rightarrow R.$$
This is proved with an argument similar to the one in A1 above - the calculation is similar.
If you change the definition of the map $\phi$ to the map
$$\psi: R\otimes_k R \rightarrow R$$
with $\psi(a\otimes b):=a\mu(\overline{b})$ you get a well defined left $R$-linear map
$$\psi_{\mu}: R\otimes_k R/I^{n+1} \rightarrow R.$$
For the composed map
$$D_{\mu}:=\phi \circ \Delta: R \rightarrow R$$
to be a differential operator, you must prove it factors via an $R$-linear map
$$\phi_1: R\otimes_k R/J^{n+1} \rightarrow R$$
with $D_{\mu}=\phi_1 \circ d^n$
where $d^n$ is the "canonical differential operator" $d^n: R \rightarrow R\otimes_k R/J^{n+1}.$
The map $E_{\mu}:= \psi_{\mu} \circ d^n$ is by the above argument a differential operator of order $\leq n$.
Jantzen defines $Dist(G,I):=\cup_u Dist^n(G,I)$ by taking the union.
Maybe the fact that $\mu$ is a differential operator helps in the case when $G$ acts on itself.
In Jantzens book he defines the distribition $Dist(G)$ using
$Dist(G,I)$ where $I$ corresponds to the unit element $e\in G$. If $k$ is a field of characteristic zero and $Lie(G)$ is the Lie algebra of $G$ there is an isomorphism
$$Dist(G) \cong U(Lie(G))$$
where $U(Lie(G))$ is the universal enveloping algebra of $Lie(G)$. There is a canonical filtration $U^n(LieG))\subseteq U(Lie(G))$ and an induced isomophism
$$Dist^n(G) \cong U^n(Lie(G)).$$
Hence the algebra of distributions generalize the universal enveloping algebra of the Lie algebra of $G$ to group schemes over Dedekind domains or more general base schemes $S$.
Another approach: If $k$ is a field and if $P^n_{G/k}:=R\otimes_k R/J^{n+1}$ it follows there is an isomorphism
$$R/I^{n+1} \cong \mathcal{O}_{G,x}/\mathfrak{m}_x^{n+1}$$
where $x\in G$ is the point corresponding to $I \subseteq R$. There is an exact sequence
$$0 \rightarrow I^{n+1}\rightarrow R \rightarrow R/I^{n+1} \rightarrow 0$$
and localizing (let $S:=R-I$) you get since $R/I^{n+1}$ is a local ring
$$S^{-1}(R/I^{n+1}) \cong R/I^{n+1} \cong S^{-1}R/S^{-1}I^{n+1}\cong \mathcal{O}_{G,x}/\mathfrak{m}_x^{n+1}.$$
There is moreover
an isomorphism
$$P^n_{G/k} \otimes \kappa(x) \cong \mathcal{O}_{G,x}/\mathfrak{m}_x^{n+1}.$$
By definition
$$Hom_R(P^n_{G/k}, R) \cong Diff^n_k(R,R)$$
which is the module of n'th order differential operators on $G$. There is an equality
$$Hom_k(R/I^{n+1}, R/I) \cong Hom_{\kappa(x)}(P^n_{G/k}\otimes \kappa(x), \kappa(x)), $$
hence the space of distributions $Dist^n(G,I)$ is the dual of the fiber $P^n_{G/k}\otimes \kappa(x)$. There is an isomorphism $P^1_{G/k}\cong R \oplus \Omega^1_{G/k}$ and hence
$$P^1_{G/k}\otimes \kappa(x) \cong \kappa(x)\oplus \Omega^1_{G/k}\otimes \kappa(x).$$
The dual $T_{G/k}:=(\Omega^1_{G/k})^*$ has
$$ T_{G/k}\otimes \kappa(x) \cong Lie(G)$$
is the Lie algebra $Lie(G)$ of $G$.
Addendum: Let me add that in the Waterhouse book he defines for any affine algebraic group $G$ with coordinate ring $k[G]$, the $k$-Lie algebra $Lie(G)$ as a Lie sub algebra (of left invariant vector fields) of $Der_k(k[G])$. When $G$ is smooth it follows $Diff_k(k[G])$ is the ring of differential operators on $G$ and there is a canonical map
$$U(Lie(G)) \rightarrow Diff_k(k[G]),$$
hence when the characteristic is zero you get a canonical map
$$Dist(G) \rightarrow Diff_k(k[G]).$$
