I don't think so for both. We can just ignore the fact that there are draws. Now we draw a directed graph, with vertices represent players and edges $a\to b$ if b beats a. For every player $p$, consider the first sum minus the second sum. If we sum them all for every player, the value is 0. So it is impossible for it to be positive or negative.

I don't think that oneplusone's sum is always 0. Let $n$ be the number of players and let $G$ be the directed graph representing wins and losses. Then the player corresponding to vertex $v$ in $G$ receives $points(v)=\frac{1}{2}(n-1+deg^+(v)-deg^-(v))$ points. Now oneplusone's sum $S$ is \begin{align*} S&=\sum_{v \in G}\left(\sum_{vw \in E(G)}points(w)-\sum_{wv \in E(G)} points(w)\right)\\ &=\sum_{w \in G} points(w)(deg^-(w)-deg^+(w))\\ &=\frac{n-1}{2}\sum_{w \in G}(deg^-(w)-deg^+(w)) - \frac{1}{2}\sum_{w \in G}(deg^+(w)-deg^-(w))^2\\ &=- \frac{1}{2}\sum_{w \in G}(deg^+(w)-deg^-(w))^2\\ &\leq 0 \end{align*} Therefore, this method discounts only one of possibilities (a) and (b). (Side question: is there a direct combinatorial or probabilistic way to see that the sum is never positive?)