Let $F$ be a free group. Does it follow that $F$ has the following property: for any $\{e,\circ, ^{-1}\}$-identity $s\approx t$ (in the sense of universal algebra), $F\models s\approx t$ if and only if $G\models s\approx t$ for all groups $G$?
If $F$ has the above property, does it follow that $F$ is free?
Is the answer to 1. and 2. the same for all kinds of algebraic structures: abelian groups, rings, ...?
If it is false for some, is it at least true for kinds of algebraic structures defined without axioms, such as magmas?
My motivation for these question is that I heard the vague statement that a free group is a group satisfying all equations forced by the group axioms but no other, and 1. is a way of making this idea precise. (But I'm not sured if that is meant!)
Here is what is meant:
Consider the free group/ring/... on $n$ elements with generators $y_1, ..., y_n$. Consider any potential identity $t \approx q$ where $t, q$ are terms build from variables $x_1, ..., x_n$ and the group/ring/... operations. Then this identity holds when substituting $y_i$ for $x_i$ in the free group/ring/... if and only if it holds for all groups/rings/...
This follows from the definition of a free algebraic structure.