It actually is the derangement problem in disguise: in each problem there is exactly one forbidden target for each element of $[n]$, and each element of $[n]$ is a forbidden target for exactly one element of $[n]$. Let $\pi$ be the following permutation of $[n]$:
$$\pi:[n]\to[n]:k\mapsto\begin{cases}
1,&\text{if }k=n\\
k+1,&\text{otherwise.}
\end{cases}$$
A permutation $\sigma$ of $[n]$ satisfies your condition if and only if $\pi\circ\sigma$ is a derangement of $[n]$. If $\Pi_n$ is the set of permutations of $[n]$, the map $\sigma\mapsto\pi\circ\sigma$ is clearly a bijection from $\Pi_n$ to itself, so $w_n=D_n$. Clearly the same is true for any $\pi\in\Pi_n$: each yields another variant of the derangement problem.
Once you have this, you know that the numbers $w_n$ satisfy the same recurrences as the derangement numbers, and there’s no real need to find a separate proof. One can prove directly that $w_n=(n-1)(w_{n-1}+w_{n-2})$ using an argument with the same basic structure as the usual combinatorial argument for the corresponding derangement recurrence, but the proof uses a somewhat more complicated version of the idea that I used above and is harder than simply exhibiting a bijection with the set of derangements.
Added: And here is such an argument. It’s quite possible that this could be simplified or cleaned up a bit: I did it fairly hurriedly.
Say that a permutation $\sigma$ of $[n]$ is good if $\sigma(k)\ne k-1$ for $k\in[n]\setminus\{1\}$, and $\sigma(1)\ne n$. Suppose that $\sigma$ is a good permutation of $[n]$. There are $n-1$ possible choices for $\sigma(n)$; suppose that $\sigma(n)=k\ne n-1$. There are now two possibilities. In what follows replace $k+1$ by $1$ if $k=n$.
Suppose first that $\sigma(k+1)\ne n-1$. Then $\sigma$ must map $[n-1]$ bijectively to $[n]\setminus\{k\}$ subject to certain restrictions. If $k=n$, $\sigma\upharpoonright[n-1]$ is simply a good permutation of $[n-1]$, and any good permutation of $[n-1]$ can be extended to a good permutation of $[n]$ that takes $n$ to $n$, so there are $w_{n-1}$ possibilities for $\sigma\upharpoonright[n-1]$. If $k\ne n$, the restrictions are as follows: $\sigma(\ell)\ne\ell-1$ for $\ell\in[n]\setminus\{1,k+1\}$, $\sigma(1)\ne n$, and $\sigma(k+1)\ne n-1$. Let
$$\varphi:[n]\setminus\{k\}\to[n-1]:\ell\mapsto\begin{cases}
k,&\text{if }\ell=n-1\\
n-1,&\text{if }\ell=n\\
\ell,&\text{otherwise;}
\end{cases}$$
then $\sigma\upharpoonright[n-1]$ satisfies those restrictions iff $\hat\sigma=\varphi\circ(\sigma\upharpoonright[n-1])$ is a good permutation of $[n-1]$. (E.g., $\sigma(k+1)=n-1$ iff $\hat\sigma(k+1)=\varphi(n-1)=k$.) Once again there are $w_{n-1}$ possible choices for $\sigma\upharpoonright[n-1]$, so there are altogether $(n-1)w_{n-1}$ good permutations $\sigma$ of $[n]$ such that $\sigma(k+1)\ne n-1$.
Now suppose that $\sigma(k+1)=n-1$. Then $\sigma$ must map $[n-1]\setminus\{k+1\}$ bijectively to $[n]\setminus\{k,n-1\}$, again subject to certain restrictions. If $k=n$, $\sigma$ maps $[n-1]\setminus\{1\}$ bijectively to $[n-2]$ in such a way that $\sigma(\ell)\ne\ell-1$ for $\ell\in[n-1]\setminus\{1\}$. Let
$$\varphi:[n-2]\to[n-1]\setminus\{1\}:\ell\mapsto\begin{cases}
n-1,&\text{if }\ell=1\\
\ell,&\text{otherwise;}
\end{cases}$$
then $\sigma\upharpoonright[n-1]\setminus\{1\}$ satisfies those restrictions iff $\hat\sigma=\sigma\circ\varphi$ is a good permutation of $[n-2]$. (E.g., $\hat\sigma(1)=n-2$ iff $\sigma(n-1)=n-2$.)
If $k\ne n$, the restrictions are that $\sigma(\ell)\ne\ell-1$ for $\ell\in[n-1]\setminus\{1,k+1\}$, and $\sigma(1)\ne n$. Let
$$\varphi:[n-2]\to[n-1]\setminus\{k+1\}:\ell\mapsto\begin{cases}
\ell,&\text{if }1\le\ell\le k\\
\ell+1,&\text{if }k+1\le\ell\le n-2
\end{cases}$$
and
$$\psi:[n]\setminus\{k,n-1\}\to[n-2]:\ell\mapsto\begin{cases}
\ell,&\text{if }1\le\ell<k\\
\ell-1,&\text{if }k<\ell<n-1\\
n-2,&\text{if }\ell=n\;,
\end{cases}$$
and let $\hat\sigma=\psi\circ\sigma\circ\varphi$; then $\sigma$ satisfies those restrictions iff $\hat\sigma$ is a good permutation of $[n-2]$. (E.g., $\hat\sigma(1)=n-2$ iff $\sigma(\varphi(1))=n$ iff $\sigma(1)=n$.) Thus, whenever $\sigma(k+1)=n-1$ there are $w_{n-2}$ ways to choose the rest of $\sigma$, so there are $(n-1)w_{n-2}$ good permutations of $[n]$ such that $\sigma(k+1)=n-1$.
Altogether, then, we have $w_n=(n-1)(w_{n-1}+w_{n-2})$.
