Often sentences that are false, are nonetheless "true almost everywhere."
Example 0. The integers do not satisfy the cancellation law:
$$\forall a\forall xy(ax = ay \rightarrow x=y)$$
This is because we can find a counterexample, namely $a=0$ (e.g. take $x=1,y=2$).
However, the way I see it, the tuple $(a,x,y)$ lives in a $3$-dimensional free module, namely $\mathbb{Z}^3$. Since the set of all counterexamples to the cancellation law forms a $2$-dimensional subspace, hence there are only "trivially-many" counterexamples, and therefore it would be reasonable to deem the cancellation law "true almost everywhere" in $\mathbb{Z}.$
Example 1. The rational numbers do not satisfy the reciprocation law:
$$\forall x\exists y(xy=1)$$
This is because we can find a counterexample, namely $x=0$. However, since $\{x \in \mathbb{Q} \mid \neg \exists y(xy=1)\}$ forms a $0$-dimensional subspace of the $1$-dimensional vector space $\mathbb{Q},$ hence it would be reasonable to deem the reciprocation law "true almost everywhere" in $\mathbb{Q}.$
Example 2. Here's a third example. Let $P$ denote the set of all points of the projective plane, and $L$ denote the set of all lines. Then for any two points $p,q \in P$, there is a unique line $p \vee q \in L$ that passes through both points. And for any two lines in the projective $l,m$, there is a unique point $lm$ at which they intersect. This defines two binary functions, $$\vee : P \times P \rightarrow L, \quad \times : L \times L \rightarrow P.$$
They satisfy the following interesting-looking identities.
$$\frac{\forall p \in P \quad \forall r,s \in P}{(p\vee r)(p\vee s) = p}\qquad \frac{\forall l \in L\quad \forall m,n \in L}{lm\vee ln = l}$$
But none of this is quite right. Observe that for no point $p$ is $p\vee p$ is ever a well-defined line, and for no line $l$ is $ll$ ever a well-defined point. So those aren't functions we've just described, they're merely partial functions. More fundamentally, the above identities are wrong. For what if the line $p\vee r$ equals the line $p\vee s$? Then their intersection $(p\vee r)(p\vee s)$ isn't even well-defined, so it surely would not be correct to say that it equals $p.$
The point, though, is that there is only a "trivial subspace" of counterexamples to the totality of our two would-be functions, and only a trivial subspace of counterexamples to our identities.
Question. Suppose we're interested in alternative semantics for first-order logic (and/or fragments thereof) in which functions are "almost-everywhere total," and universal quantifiers are interpreted "almost-everywhere." Alternatively, suppose we're interested in adjoining a third quantifier $\forall^*$ to first-order logic, such that $\forall^*x\varphi$ is intended to assert that $\varphi$ is true for almost-all $x$. Where can we learn about these ideas?
You seem to concentrate on algebraic properties that hold on a dense open set in the sense of Zariski topology. So the latter can be considered a valid interpretation of "almost everywhere" in this context. Then again, by the very definition of Zariski topology this just means that the set of exceptions can be desribed as the zero set of a nonzero polynomial; this might look circular (the argumentation, not the zero set).