You have obviously thought deeply about this problem. I hope my answer here provides some lines of thought that you have not already come across. Apologies if not.
We may place the standard $2$-adic metric on $\mathbb{F}_2^n$. That is, if two sequences of $0$'s and $1$'s agree in the first $k$ places, but not any further, then the distance between them is $\frac1{2^k}$. Let $G_n$ denote the group of isometries of this metric space.
The actions of both $(\mathbb{Z}/2^n\mathbb{Z})^\times$ and $(\mathbb{Z}[t]/\langle t^n\rangle)^\times$ on $\mathbb{F}_2^n$ are isometries. Two see this, note that when we make the natural identification of either ring with $\mathbb{F}_2^n$, the distance between two elements is determined by the greatest power of $2$ (in $\mathbb{Z}/2^n\mathbb{Z}$) or $t$ (in $\mathbb{Z}[t]/\langle t^n\rangle$) dividing the difference. In both cases this power is not altered by multiplication by a unit.
Let $H_n=\langle (\mathbb{Z}/2^n\mathbb{Z})^\times\cup (\mathbb{Z}[t]/\langle t^n\rangle)^\times\rangle\subset {\rm Sym} (\mathbb{F}_2^n)$ be the group we are seeking to understand. We have shown that $$H_n\subseteq G_n.$$
The groups $G_n$ can be understood in a number of ways. For example inductively: $$G_{n+1}=(G_n\times G_n)\rtimes C_2,$$
where the conjugation action of $C_2$ is swapping the two factors of $G_n$. To see this note that we have a homomorphism $G_{n+1}\to C_2$ sending an isometry to its action on the first digit. The kernel of this map consists of ordered pairs: an isometry on the set of sequences beginning with $0$, and an isometry on the set of sequences beginning with $1$.
Thus we know that $|G_n|=2^{2^n-1}$. Another approach is to consider the homomorphism $G_{n+1}\to G_n$ given by 'forgetting' the action on the last digit. The kernel of this homomorphism is all isometries, which fix the first $n$ digits. For each $n$ digit sequence, we can either swap the two $n+1$ digit extensions, or not, so this kernel is $C_2^{2^n}$. Thus we have the short exact sequence: $$C_2^{2^n}\to G_{n+1}\to G_n.$$
Note as a sanity check we have: $|C_2^{2^n}|\cdot|G_n|=2^{2^n}\cdot 2^{2^n-1}=2^{2^{n+1}-1}=|G_{n+1}|$.
Now let us return to the main question - what is $H_n$?
We know $H_n$ is contained in an index $4$ subgroup of $G_n$: namely the index $4$ subgroup which either fixes the first $2$ digits, or acts as multiplication by $3\in (\mathbb{Z}/2^n\mathbb{Z})^\times $ (or equivalently $1+t\in \mathbb{Z}[t]/\langle t^n\rangle$) on them.
Thus, in particular we know that $|H_n|=2^m$, with $m\leq 2^n-3$.
Further, we know that $H_n$ acts separately on strings starting with a $0$ and strings starting with a $1$. Thus, forgetting the action on strings starting with a $0$, we get a homomorphism: $$f\colon H_n\to G_{n-1}.$$
Question 1 Is the homomorphism $f$ surjective?
Question 2 Is the homomorphism $f$ injective?
If the answer to both questions is yes, then we have identified $H_n$ as precisely $G_{n-1}$. For Question 2, the answer being yes corresponds to the action of an element of $H_n$ being completely determined by its action on strings beginning with a $1$. The only evidence we can offer for this is that it holds for elements of both $(\mathbb{Z}/2^n\mathbb{Z})^\times$ and $(\mathbb{Z}[t]/\langle t^n\rangle)^\times$. Of course this does not imply that it holds for compositions of such elements.
For Question 1 we can say a little more. We can verify that the answer is yes up to $n=3$, and make an attempt at an inductive proof that the answer is yes. Note that the answer to Question 1 being yes would imply that $H_n\subset G_n$ and $H_n$ has $G_{n-1}$ as a quotient, effectively trapping it between two known groups.
$H_1$ is the trivial group. $H_2$ is generated by the swapping of the two sequences beginning with $1$ as this coincides with both $(\mathbb{Z}/2^2\mathbb{Z})^\times$ and $(\mathbb{Z}[t]/\langle t^2\rangle)^\times$ acting on $\mathbb{F}_2^2$. Thus the map $f$ is surjective for $n=2$.
For $n=3$ the inverse of the action of $(1+t)$ composed with the action of $3$ fixes all sequences of length $3$, except for $101$ and $100$, which it swaps.
Conjugating this swap by the action of $3$ (or $1+t$) we can swap only the other pair of sequences which begin with a $1$.
As we already know that $f$ is surjective for $n=2$, we know that elements of $H_3$ can induce any element of $G_2$ up to action on the final digit (that is swapping pairs of sequence which begin with a $1$ and differ only in the final digit). However we have now shown that $H_3$ also contains these actions on the final digit. Thus $f$ is surjective for $n=3$.
Finally we attempt to generalise this to an inductive proof that $f$ is always surjective. Suppose that $f$ is surjective for some strings of length $n-1$. Then we know that $H_n$ can induce any element of $G_{n-1}$ on strings beginning with a $1$, up to swapping some final digits. Thus to show that $f$ is surjective on strings of length $n$, we must show that $H_n$ contains an element that fixes the first $n-1$ digits of strings beginning with a $1$, and flips the final digit on any desired subset of pairs of strings that start with a $1$ and agree on the first $n-1$ digits.
Firstly consider the actions of $x=(1+t^{n-2})$, and with $y=(1+2^{n-2})$. These actions only differ in the final digit, on strings beginning with a $1$ and whose second to last digit is a $1$. Thus $xy^{-1}\in H_n$ fixes the first $n-1$ digits of all sequences beginning with a $1$, and flips the final digit of all such sequences which have a $0$ as second to last digit.
By the inductive hypothesis, we have an element $z\in H_n$ which flips the second to last digit of the four sequences beginning $1$ followed by any specified sequence $\alpha$ of $n-3$ digits, and fixes all other strings.
Then $z(xy^{-1})z^{-1}(xy^{-1})\in H_n$ will flip the final digit on the two pairs of strings beginning $1$ followed by $\alpha$ (as all the other flips are common to both $z(xy^{-1})z^{-1}$ and $(xy^{-1})$, hence cancel). Unfortunately this is as far as I can get. If we could flip the final digits of any single pair of sequences beginning with $1$ and agreeing on the first $n-1$ digits, then we would have completed the inductive step. However I have only been able to show that this can be done $2$ at a time - for a pair of such pairs, differing in the second to last digit.
Proof that $f$ is injective:
Let $L$ denote the shift operator on sequences in $\mathbb{F}_2^n$, corresponding to multiplication by $2$ and multiplication by $t$. As $L$ commutes with all elements of $(\mathbb{Z}/2^n\mathbb{Z})^\times$ and all elements of $(\mathbb{Z}[t]/\langle t^n\rangle)^\times$, we know that $L$ commutes with all elements of $H_n$.
Suppose $f(g)=f(g')$. Given $x\in \mathbb{F}_2^n$, we have $x=L^ru$, where the first digit of $u$ is $1$. Thus $$gx=gL^ru=L^rgu=L^rg'u=g'L^ru=g'x,$$ so $g=g'$.
From comments below by @AlekseiAverchenko we know $f$ is not surjective in general. In particular, not for $n=4$:
We have $|H_1|=2^0, |H_2|=2^1, |H_3|=2^3, |H_4|=2^6$.
For $n=1,2,3$ we have $H_n\cong G_{n-1}$. However $|G_3|=2^7$, so $H_4\not\cong G_3$. In general $H_n\subseteq G_{n-1}$.
