A good way to think about these equations is to imagine them as sets over valid words. Say I have an alphabet $\Sigma = \{a, b\}$, then the rule
$$
A \leftarrow A a \mid b
$$
says that there's some set $A$ that is isomorphic to the set $(A * \{a\}) \cup \{b\}$. Here the $*$ operation is the product/concatenation operation lifted over sets, that is
$$
\{a, b, c\} * \{d, e, f\} = \{ad, ae, af, bd, be, bf, cd, ce, cf\}
$$
Going back to our grammar rule, it is basically declaring the existence of a set $A$ of words that satisfies the equation
$$
A = (A * \{a\}) \cup \{b\}.
$$
Now, there's a huge class of sets of words that satisfies this equation. For example, the set $A_\bot = \{b, ba, baa, baaa, baaaa, \dots\}$ is a valid solution since a $b$ or appending any element of $A_\bot$ with another $a$ will result in a word that is already in $A_\bot$. However, you should check that
$$
A_{bb} = \{b, bb, ba, bba, baa, bbaa, baaa, bbaaa, \dots\}
$$
is also a valid solution to this equation. In particular, $b$ is already in $A_{bb}$, and for any element $ba^k$ or $bba^k$ in $A_{bb}$, its concatenation with $a$ results in $ba^{k+1}$ and $bba^{k+1}$, which are also already in $A_{bb}$. Therefore, $A_{bb}$ is also a valid fixed point.
In fact, for any given ground set of words $S = \{w_1, w_2, w_3, \dots\}$, we can construct a closure $A_S$ which contains every word in $S$ and similarly satisfies the above equation. In particular, the closure operator is the same isomorphism:
\begin{align*}
A_S^{k + 1} &= (A_S^k * \{a\}) \cup \{b\} \\
A_S &= \bigcup_k^\infty A_S^k
\end{align*}
So now we get to the real question. If there are an infinite supply of solutions for this equation, then what good is it as a characterization of some grammar? Well, intuitively, we hope to just characterize those words that are generated from this rule "from scratch". In effect, we wish to treat these rules as free objects. That is, we care about the case where the underlying generator $S = \{\}$. Let
$$
A(S) = (S * \{a\}) \cup \{b\}
$$
then we want
$$
A_* = \bigcup_k A^{(k)}(\{\})
$$
to be the set generated by our grammar rule, where $A^{(k)} = A(A(\stackrel{k}{\dots} A(\cdot)))$.
This then gives some intuition of what we mean by the "least solution." $A(S)$ is monotone in the sense that $S \subseteq S' \implies A(S) \subseteq A(S')$; in fact, this holds over all grammar rules. Since the least element over the class of possible sets that $S$ can take (ordered on set inclusion) is the empty set, then effectively the "freely generated" language that we desire also turns out to be the smallest (ordered on set inclusion) fixed point, that is, $A(\{\}) \subseteq A(S')$ for any set $S'$.
In general, you'll see a lot of these "smallest solution" clauses all around the landscape of computer science. For example, any recursive program is the smallest solution to some program equation; any constructable inductive datatype is the smallest solution to some inductive datatype equation; any valid program semantic is the smallest solution to some logical equation; etc. It turns out that these classes of topological closures corresponds elegantly to a notion of finiteness. For example, the set of all finitary words $\Sigma^*$ is itself a least fixed-point. Since computations themselves can be seen as some enumeration process of finite objects, the analogy holds quite naturally; least fixed-points typically gives some assurance of some form of computability.
In conclusion, if you see the phrase "least solution" or "the least fix-point," it often means that the problem concerns finite objects that are freely-generated in the sense that only things that can be derived "from scratch" are considered.
