Background and Notation
A single Nim heap of size $n$ will be denoted "$*n$", with $*1$ abbreviated "$*$" and $*0$ abbreviated "$0$".
"$+$" denotes the disjunctive sum, so that $*+*3$ is a Nim position with one heap of size $1$ and one heap of size $3$.
This is not standard, but we can take a page from Topology and call a game $G$ connected if it is not isomorphic to a disjunctive sum other than $G+0$ and $0+G$.
Edit: "$\vee$" denotes a variant of the selective sum, so that when $G$ and $H$ are nim heaps, $G\vee H$ is a game in which a player can move in $G$ or $H$ or both simultaneously but if both then only one token/chip is removed from each.
The game described by user89 is one in which there are two Nim heaps (one being $*$), and a player can move in one or both of them on their turn. This means that this game with $n+1$ chips is $*n\vee*$.
Is $*n\vee*$ connected for each $n$?
Nim heaps are connected
$0$ and $*$ are connected since a nontrivial disjunctive sum must have a line of play (a "run") of length at least $2$, since each summand must have at least one move.
For $n\ge2$, $*n$ is connected since it has a move to $0$, but a disjunctive sum $G+H$ does not if neither $G$ nor $H$ are $0$.
The OP's game is connected
Note that $0\vee*$ is isomorphic to $*$, which is connected.
Note that $*\vee*$ is isomorphic to $*2$, which is connected.
If $n\ge2$, then $*n\vee*$ has a move to $0\vee*$ and a move to $*\vee*$. Suppose $*n\vee*=G+H$ with $G,H$ nonzero. Since there is a move to $*$, then one of $G$ and $H$ must be $*$ since otherwise there is no run of length $2$. Suppose $H$ is $*$. But, then, since there is a move to $*2$, and $*2$ is connected, we $G$ must be $*2$. For $n>2$ we have a contradiction since $*n\vee*$ has a longer run than any run of $*2+*$.
For $n=2$, we have a contradiction since $*2+*$ has a move to $*+*$, but $*2\vee*$ only has moves to $*2$ (two of them) and to $*$.
Suppose, for sake of induction, that for some $k\ge 1$ we have shown that $*n\vee*k$ is connected. Then if we have a position like $*m\vee*(k+1)$, it has moves to $*(k+1)$ and to $*m\vee*k$, both of which are connected, distinct (unless $m=1$ and $k=1$, which was handled earlier) and nonzero. Then if $*m\vee*(k+1)$ is a disjunctive sum, it must be the sum of those two. But the length of the longest run in the disjunctive sum is $m+2k+1$, and the length of the longest run in $*m\vee*(k+1)$ is the smaller $m+k+1$.