I am looking at the plethysm "negation rule,"
Theorem 6
If $g\in\Lambda^n$ is homogeneous of degree $n$ and $A$ is any plethystic alphabet, then $$g[-A]=(-1)^n\big(\omega(g)\big)[A].$$
In particular, we should have $p_k[-A]=-p_k$. However, I am running into trouble with specific examples, such as \begin{align} p_2[-p_1]&=p_2\left[-\sum_i x_i\right]\\ &=p_2\left[\sum_i (-x_i)\right]\\ &=p_2(-x_1,-x_2,\ldots)\\ &=\sum_i(-x_i)^2\\ &=\sum_i x_i^2\\ &=p_2[p_1] \end{align}
Clearly I have some misunderstanding with how to compute plethysm. I think my issue in the above example is the equality $p_2\left[\sum_i (-x_i)\right]=p_2(-x_1,-x_2,\ldots)$ since the linked article only mentions a "monomial substitution rule" for a finite sum of monomials:
Theorem 7
Suppose $A$ is a finite sum of monic monomials $M_1,\ldots,M_N$ in $Z$. For any $g\in\Lambda$, $$g[A]=g(M_1,M_2,\ldots,M_N),$$ [...]
However, Stanley's definition of plethysm in Enumerative Combinatorics, Vol. 2 seems to allow any countable sum of monomials:
[EC2, Stanley]
Suppose that the symmetric function $f\in\Lambda$ is a sum of monomials, say, $f=\sum_{i\geq1}x^{a^i}$. Given $g\in\Lambda$, define the plethysm $g[f]$ by $$g[f]=g(x^{a^1},x^{a^2},\ldots).$$
What is the error in my computation? Does the "monomial substitution rule" work for countably infinite sums?
Your mistake is coming from your definition of "monomial". Whenever monomials are mentioned in the context of symmetric functions, the author usually means "monic monomials" (i.e. a monomial with coefficient $1$). So, in your example, $-p_k$ is in fact not a sum of monomials because the term $-x_i$ is not a monic monomial (it has a coefficient of $-1$). In fact, this condition is explicitly stated in the "Theorem 7" you cited.