Thanks for this interesting question. From your Proposition 1.2 it would seem that the non-universal positive integers contains the set
$$
\{k\in\mathbb{Z}_{\geq 0}: \exists p\leq k\mbox{ such that } \prod_{\substack{i=0\\k_i>0}}^r (p-k_i) <p-1\}
$$
where
$$k=\sum_{i=0}^rk_ip^i$$ and $p$ is prime.

Let $x=p^{m+1}$. Let $p$ be prime and $n=\sum_{i=0}^m n_ip^i$ be the expansion of $n$ in base $p$ where $0\leq n_i<p$. Let
$$
\tau_k(n)=\sum_{\substack{(d_1,\dots,d_k)\in\mathbb{Z}^k\\1\leq d_1,\dots,d_k\leq n\\d_1\dots d_k=n}}1.
$$
Let $$A_{p,m}=\{n\in\mathbb{Z}: p < n = \sum_{i=0}^m n_ip^i < x, \prod_{\substack{i=0\\n_i>0}}^m(p-n_i)<p-1 \}.$$
Then
\begin{eqnarray*}&&|A_{p,m}|\\
&=&\sum_{j=1}^{p-2}|\{n\in\mathbb{Z}: p < n = \sum_{i=0}^m n_ip^i < x, \prod_{\substack{i=0\\n_i>0}}^m(p-n_i)=j \}|\\
&=&\sum_{j=1}^{p-2}|\{n\in\mathbb{Z}: p \leq n = \sum_{i=0}^m n_ip^i < x, \prod_{\substack{i=0\\n_i>0}}^m(p-n_i)=j \}|\\
&=&\left(\sum_{j=1}^{p-2}|\{n\in\mathbb{Z}: 0 \leq n = \sum_{i=0}^m n_ip^i < x, \prod_{\substack{i=0\\n_i>0}}^m(p-n_i)=j \}|\right)-(p-1)\\
&=&\left(\sum_{j=1}^{p-2}\sum_{k=1}^{m+1}{m+1 \choose k}\tau_k(j)\right) + 1 - (p-1)
\end{eqnarray*}
as
$\prod_{\substack{i=0\\n_i>0}}^m(p-n_i)=1$
when $n=0,p-1$ and $\prod_{\substack{i=0\\n_i>0}}^m(p-n_i)=p-n$ when $1< n < p-1$ and furthermore, the number of ways to distribute $m+1$ digits $n_i\in\{0,\dots,p-1\}$ such that $\prod_{\substack{i=0\\n_i>0}}^m(p-n_i)=j$ where the fixed $j$ satisfies $1 \leq j < p-1$ and $k$ of the $m+1$ digits $n_i$ are in $\{2,\dots,p-1\}$, is ${m+1 \choose k}\tau_k(j)$.

It is known that (https://terrytao.wordpress.com/2016/08/31/heuristic-computation-of-correlations-of-higher-order-divisor-functions/)
$$
\sum_{n \leq y}\tau_k(n) \sim \frac{\log^{k-1}y}{(k-1)!}y
$$
and hence
\begin{eqnarray*}
|A_{p,m}| &=& \left(\sum_{k=1}^{m+1}{m+1 \choose k}\sum_{j=1}^{p-2}\tau_k(j)\right) - (p-2)\\
&\ll&\sum_{k=1}^{m+1}{m+1 \choose k}\frac{\log^{k-1}p}{(k-1)!}p\\
&\ll&2^{m+1}p\log^m(p)
\end{eqnarray*}
as $p \rightarrow \infty$ for fixed $m$. Hence if $x=p^{m+1}$, $|A_{p,m}|/x \rightarrow 0$ as $p\rightarrow\infty$ for fixed $m$.

On the other hand, fix $x$ and let
$$B_{p,x}=\{n\in\mathbb{Z}: p < n = \sum_{i=0}^m n_ip^i < x, \prod_{\substack{i=0\\n_i>0}}^m(p-n_i)<p-1 \}.$$
Then it is less clear what
$$
\lim_{x \rightarrow \infty}\frac{|\cup_{p<x}B_{p,x}|}{x}
$$
is.