As in your first example, you have two (not one) free variable. You let $x= s$, $y = t$ in the first example and $z = t$ follow from the equation. In this example you must take $x$ but you can choose between $y$ and $z$ to be free for you.
In the second example you can take any two variables as free. Let's take $x = s$, $y = t$. The equation $x-2y+5z = 0$ gives
$$ z = -\frac 15 x + \frac 25y = \frac 15(2t -s) $$
So a basis is given by $(1,0, -\frac 15)^\top$, $(0,1,\frac 25)^\top$.
You must note that $A$ is a vector in $\Bbb R^2$ and $A'$ is a vector in $\Bbb R^3$. That being said, the span of $A$ is not the same as the span of $A'$ because they live in different spaces!
However, there is some sense in which these spans are the "same." Consider that for any vector $v \in \operatorname{span}A$, $v = \lambda(a, a)$ for some $\lambda \in \Bbb R$. Similarly, for any vector $u \in \operatorname{span}A'$, $u = \mu(a, a, a)$ for some $\mu \in \Bbb R$. Both vectors, though they live in different spaces, effectively encode one piece of data, namely the number $\lambda a$ or $\mu a$.
More precisely, we say there is an isomorphism, or a relabeling of these spaces with the set of real numbers, $\Bbb R$. That is, for any real number $r$, there is a vector in $\operatorname{span} A$ and $\operatorname{span} A'$ such that the components of that vector are each the number $r$. Similarly, given a vector in $\operatorname{span} A$ or $\operatorname{span} A'$, we can find a real number that is equal to that vector's single piece of data.
It is not the case that the span of any vector in $\Bbb R^2$ or $\Bbb R^3$ is equal to $\Bbb R$, but it is the case that they can be isomorphic to $\Bbb R$.
In symbols, we write
$$
\operatorname{span}A \cong \operatorname{span}A' \cong \Bbb R.
$$
To address your main question, the definition of a vector space is a set closed under an addition $+$ and scalar multiplication $\cdot$ such that various other conditions hold. The glaring issue with your saying that $\operatorname{span}A$ and $\operatorname{span}A'$ are equal is that we cannot add vectors in $\operatorname{span}A$ and $\operatorname{span}A'$, so this does not respect closure under addition.
Since not both of your premises are true, the conclusion that $\operatorname{span}A$ and $\operatorname{span}A'$ are equal is invalid. The conclusion that they can be isomorphic is valid, as I have described.
Best Answer
If I understand the question, you want a spanning set for the $xy$-plane in ${\mathbb R^3}$. A spanning set for the $xy$-plane is $${\mathcal{B}}=\left\{\begin{bmatrix}a\\b\\0\end{bmatrix},\,\begin{bmatrix}c\\d\\0\end{bmatrix}\right\}, $$ where $ad-bc\neq 0$. Any choice of $a,\,b\,c$ and $d$ which satitfy the condition are a spanning set.
Some examples: $$\begin{align}\mathcal{B}_1&=\left\{\begin{bmatrix}1\\0\\0\end{bmatrix}, \,\begin{bmatrix}0\\1\\0\end{bmatrix}\right\}&\text{[the standard basis]}\\ \mathcal{B}_2&=\left\{\begin{bmatrix}1\\1\\0\end{bmatrix}, \,\begin{bmatrix}1\\-1\\0\end{bmatrix}\right\}\\ \mathcal{B}_3&=\left\{\begin{bmatrix}1\\-4\\0\end{bmatrix}, \,\begin{bmatrix}2\\1\\0\end{bmatrix}\right\} \end{align} $$