There exist algebraic structures $X$ with no minimal generating set. For example, $\mathbb{Q},$ viewed as an Abelian group. There also exist algebraic structures whose every generating set includes some minimal generating set. e.g. $\mathbb{Q}$, viewed as a $\mathbb{Q}$-module.
Question. Are there reasonably general conditions under which we can expect an algebraic structure $X$ to have the property that for every generating set $G \subseteq X,$ there is a minimal generating set $M \subseteq G$?
I do not know about the situation in the generality you describe, but here is one special case that may be of interest to you. In the following, all rings are commutative because I am a commutative ring theorist and I don't remember what carries over to non-commutative rings.
For a ring R the following are equivalent:
1. $R$ is quasilocal and every $R$-module has a minimal generating set.
2. R is a connected Bass ring, i.e., every nonzero R-module has a maximal submodule.
3. R is a Steinitz ring, i.e., the conclusion of the Steinitz Exchange Lemma holds for every free R-module.
4. R is a connected perfect ring, i.e. it satisfies the DCC on principal ideals or equivalently every R-module has a projective cover.
5. For every sequence $x_1,x_2,\ldots$ of nonunits, there is an $n$ with $x_1\cdots x_n = 0$.
In more generality, it is known that any perfect ring is a ring over which modules have minimal generating sets, but the converse is false. However, any ring with the stronger property that every generating set contains a minimal generating set must be perfect. I am not sure about the converse. There is a paper "A characterization of left perfect rings" that claims that the converse is true, but there is a mistake in the proof. (Edit: Sloppy wording.)