I recently managed to convert this problem into a cellular automata, and the answer to the second question appears to be no, making this question uninteresting.
However, I think some people might appreciate the details of the computations, so I'll put them here (and then close the question? I'm not sure how I should deal with answering my own question.)
First, the results of the computations. Let $N(x)$ denote the number of values $n$ below $x$ such that $n$ copies of $h$ and $n+1$ copies of $h$ have the same outcome. Then
$N(10^3) = 15, N(10^6) = 86, N(10^9) = 148, N(10^{12}) = 196,N(10^{15}) = 205,$ and $N(10^{54}) = 205$. The last value of $n$ below $10^{54}$ is (if I didn't make any mistakes) $3,814,460,171,075$.
Next, the details of the computation. I start by visualizing the situation as a coloring of $\mathbb{N}^3$ in the colors red, green, and blue. The coloring of $(i,j,k)$ corresponds to the outcome of the game formed by $i$ copies of $h$, $j$ copies of $g$, and $k$ copies of $*2$ as follows: green corresponds to a lost game, red corresponds to a game that loses when you add $*$, and blue corresponds to a game that is a first player win and remains a first player win when you add $*$ to it. In order to calculate the color of $(i,j,k)$, one needs to know the colors of $(i,j,k-1)$, $(i,j-1,k)$, $(i,j-1,k+2)$, $(i,j-1,k+3)$, $(i-1,j,k+1)$, $(i-1,j+1,k)$, and $(i-1,j+2,k)$.
In order to convert this to a cellular automata, I take "diagonal cross sections" of $\mathbb{N}^3$. I lay out one such diagonal cross section as follows:
(3,0,0) (3,0,1) (3,0,2) (3,0,3) (3,0,4) (3,0,5) (3,0,6) (2,0,7) (2,0,8) ...
(3,1,0) (3,1,1) (2,1,2) (2,1,3) (2,1,4) (2,1,5) (2,1,6) (2,1,7) ...
(2,2,0) (2,2,1) (2,2,2) (2,2,3) (2,2,4) (2,2,5) (2,2,6) ...
(2,3,0) (2,3,1) (2,3,2) (2,3,3) (1,3,4) (1,3,5) ...
(1,4,0) (1,4,1) (1,4,2) (1,4,3) (1,4,4) ...
...
To go from this cross section to the next higher cross section, we just need to replace the cell marked $(2,0,7)$ with $(3,0,7)$, the cell marked $(2,1,2)$ with $(3,1,2)$, the cell marked $(1,3,4)$ with $(2,3,4)$, etc. With the cells laid out like this, it is easy to see that these updates only require the knowledge of the colors of cells within two steps of the cell you would like to update. Unfortunately, most cellular automata only allow the transition rules to depend on the values of the cells immediately adjacent, so I had to emulate this cellular automata with a slightly more complicated one.
The end result is this rule table, with these colors which you can load in a program such as Golly and experiment with. The automata needs a single white square to get started, and then proceeds to compute winning positions automatically. When it detects an $n$ such that the outcome of $(n,0,0)$ matches the outcome of $(n+1,0,0)$, it fires a cyan signal upwards, so you can calculate $N(x)$ by simulating up to generation $12x+14$ and counting the number of cyan signals produced. Here is a screenshot of a computation in progress:
Best Answer
Let $\oplus$ denote the bitwise xor operation on natural numbers. It is both well-known and easy to verify that a Nim position $(n_1,\dots,n_k)$ is a second player win in misère Nim if and only if some $n_i>1$ and $n_1\oplus\cdots\oplus n_k=0$, or all $n_i\le1$ and $n_1\oplus\cdots\oplus n_k=1$.
If I understand it correctly, this translates to the following structure. Let $A=(\omega,\oplus,0)$ (in other words, $A$ is the direct sum of countably many copies of the two-element abelian group), let $A_0=\{0,1\}$ be its submonoid, and let $B=(\{0,1\},\lor,0)$ be the two-element semilattice. Then the underlying monoid of the misère quotient of Nim is the submonoid $Q=(A\times\{1\})\cup(A_0\times\{0\})$ of the product monoid $A\times B$, the misère quotient itself being $(Q,\{(0,1),(1,0)\})$. If $(n_1,\dots,n_k)$ is a Nim position, its value in $Q$ is $(n_1\oplus\cdots\oplus n_k,u)$, where $u=0$ if $n_i\in\{0,1\}$ for each $i$, and $u=1$ otherwise. $Q$ has (as a commutative monoid) the infinite presentation $\langle \{a_n:n\in\omega\},b\mid a_n+a_n=b+b=0,a_n+b=a_n+a_0\rangle$. Finitely generated submonoids of $Q$ are finite, hence $Q$ itself is not finitely generated.
(Note that the monoid corresponding to normal play Nim is just $A$.)