Let $(X, \mathcal{O}_X)$ be a locally ringed space. As a sanity check for myself, I'd like to show that the addition and multiplication of the germs at a point $p \in X$ are well-defined. I was able to show this for addition as follows:
Let $(f, U)$, $(g, V)$, $(\widetilde{f}, \widetilde{U})$, and $(\widetilde{g}, \widetilde{V})$ be germs in the stalk $\mathcal{O}_p$ such that $(f, U) \sim (\widetilde{f}, \widetilde{U})$ and $(g, V) \sim (\widetilde{g}, \widetilde{V})$. It follows that there exist open sets $W \subset U \cap \widetilde{U}$ and $W' \subset V \cap \widetilde{V}$ such that $f\vert_W = \widetilde{f}\vert_W$ and $g\vert_{W'} = \widetilde{g}\vert_{W'}$. We wish to show that $(f+g, U \cap V) \sim (\widetilde{f} + \widetilde{g}, \widetilde{U} \cap \widetilde{V})$. We begin by showing that $(f+g, U \cap V) \sim (\widetilde{f}+g, \widetilde{U}\cap V)$, which amounts to finding an open set $Z \subset U \cap \widetilde{U} \cap V$ such that $(f+g)\vert_Z = (\widetilde{f} + g)\vert_Z$. Since restriction maps are ring morphisms, this is equivalent to finding such a $Z$ where $f\vert_Z + g\vert_Z = \widetilde{f}\vert_Z + g\vert_Z$ iff $f\vert_Z = \widetilde{f}\vert_Z$. Letting $Z = W \cap V$, the result follows because $Z \subset W$ and $f$ and $\widetilde{f}$ already agree on $W$. The equivalence $(\widetilde{f}+g, \widetilde{U}\cap V) \sim (\widetilde{f}+\widetilde{g}, \widetilde{U}\cap \widetilde{V})$ is similar by letting $Z = \widetilde{U} \cap W'$. We conclude that $(f+g, U \cap V) \sim (\widetilde{f}+g, \widetilde{U}\cap V) \sim (\widetilde{f}+\widetilde{g}, \widetilde{U}\cap \widetilde{V})$, and we're done.
If I were to proceed analogously for the case of multiplication, I would not be able to use cancellation as I did above, since $\mathcal{O}_p$ is not an integral domain in general. What am I missing?
You don't need to cancel anything. You needed to show that $f\vert_Z + g\vert_Z = \widetilde{f}\vert_Z + g\vert_Z $ - and this is what you've obtained for, as you said, $f$ and $\tilde{f}$ already agree on $W$. You can just take $Z =W$ then, actually.
But you make it unnecessary complicated in general. $f$ equals $\tilde{f}$ on all opens inside $U \cap \tilde{U},$ and $g$ equals $\tilde{g}$ on all opens inside $V \cap \tilde{V}.$ Then obviously $(f,g) = (\tilde{f}, \tilde{g})$ on all opens inside $W: = U \cap \tilde{U}\cap V \cap \tilde{V},$ so any binary map (like addition) will give the same result on these pairs over $W$.