Given the following definition:
Let $M$ be a move set and $A,B \subset M^\omega$ are disjoint, we define the winning conditions for the game $G(A,B)$ as follows: if the play of the game is $x \in M^\omega$, then player I wins if $x \in A$, player II wins if $x \in B$, and otherwise the game is a draw. A strategy $\sigma$ is winning or drawing in $G(A,B)$ for player I if for all strategies $\tau$ , we have that $\sigma * \tau \in A$ or $\sigma * \tau \notin B$, respectively. A strategy $\tau$ is winning or drawing in $G(A,B)$ for player II if for all strategies $\sigma$, we have that $\sigma * \tau \in B$ or $\sigma * \tau \notin A$, respectively.
what would a "drawing strategy in $G(A, B)$" mean? A strategy such that a player can at least play a draw (so either draw or win), or a strategy such that a player can force a draw (so the game always ends in a draw). I find the formulation of the definition a bit unclear in this regard (but maybe that's just me?), and I think these two options are not equivalent.
A drawing strategy is one which either draws or wins against every strategy - or perhaps more snappily, one which never loses. The point (looking at drawing strategies for player $I$ for simplicity) is that $\sigma*\tau\not\in B$ means that player $II$ does not win the play of $\sigma$ against $\tau$ - that is, that $\sigma$ either draws against $\tau$ or wins against $\tau$.
(In particular, $\sigma*\tau\in A$ does imply, but is not implied by, $\sigma*\tau\not\in B$.)