No it does not mean that adding one root gives you all the roots. For example let us calculate the splitting field of $x^4 - 6 \in \Bbb{Q}[x]$. By Eisenstein's Criterion with $p= 2$ this polynomial is irreducible over $\Bbb{Q}$. Now we find the roots of this polynomial in $\Bbb{C}$ to be
$\sqrt[4]{6}e^{2 k \pi i/3} \hspace{2mm} \text{for} \hspace{2mm} k=0, \ldots 5.$
Now you may guess that adjoining one root, say $\sqrt[4]{6}e^{2 \pi i/3}$ is enough to give us the splitting field. But then we have a problem because
$\Bbb{Q}(\sqrt[4]{6}e^{2 \pi i/3}) \cong \Bbb{Q}[x]/(x^4 - 6) \cong \Bbb{Q}[\sqrt[4]{6}].$
The field on the right hand side only contains two of the roots of the polynomial $x^4 -6$, so there must be something wrong. In fact the splitting field is actually
$\Bbb{Q}(\sqrt[4]{6},e^{2\pi i/3}) \cong \Bbb{Q}(\sqrt[4]{6}, \sqrt{3}i)$
that has degree 8 over $\Bbb{Q}$. It is a good exercise which you should do to show why this is the splitting field.
In some polynomial rings over certain fields, adjoining one root to the field does give you all the roots. Consider the polynomial
$f(x) = x^p - x - a$
in $\big(\Bbb{Z}/p\Bbb{Z}\big)[x]$ with $a \neq 0$. Then in $\Bbb{Z}/p\Bbb{Z}$, no matter what value we substitute in for $x$ by Fermat's Little Theorem $f(b) = a$ for all $b \in \big(\Bbb{Z}/p\Bbb{Z}\big)$. Now create a root for this polynomial in the field extension
$\big(\Bbb{Z}/p\Bbb{Z}[x]\big)/(x^p - x - a).$
Call that root $\gamma$. Now we know that this field must contain $\big(\Bbb{Z}/p\Bbb{Z}\big)$ because
$\big(\Bbb{Z}/p\Bbb{Z}\big) \subset \big(\Bbb{Z}/p\Bbb{Z}[\gamma]\big) \cong \big(\Bbb{Z}/p\Bbb{Z}[x]\big)/(x^p - x - a).$
Now we can view the polynomial $x^p - x - a$ as sitting inside $\big(\Bbb{Z}/p\Bbb{Z}[\gamma]\big)[x].$ Furthermore, this is a polynomial ring over a field because the element $\gamma$ is algebraic over $\Bbb{Z}/p\Bbb{Z}$. We know any polynomial in a polynomial ring over a field can have at most $\deg f$ roots. In your case if we find $\deg f$ distinct roots for the polynomial $f(x) = x^p - x - a$ lying in that field, we must have found all of them. The roots of $f(x)$ are distinct because the greatest common divisor of $f(x)$ and its formal derivative $f'(X) = px^{p-1} - 1 = -1$ is 1.
Now notice for any $k \in \Bbb{Z}/p\Bbb{Z}$,
$\begin{eqnarray*} f(\gamma + k) &=& (\gamma + k)^p - \gamma - k -a\\ &=& \gamma^k + k^p - \gamma - k - a \hspace{2mm} \text{(Using the fact that $p| \binom{p}{i}$ for $1 \leq i \leq p-1$})\\ &=& \gamma^k - \gamma - a \hspace{6mm} \text{(Fermat's Little Theorem shows that $k^p - k = 0$)}\\ &=& 0 \end{eqnarray*}$
because $\gamma$ is a root of $f$. Letting $k$ vary from $0$ to $p-1$ shows that we have $p$ distinct roots for $f(x)$ lying in the field extension $\big(\Bbb{Z}/p\Bbb{Z}\big)[\gamma]$. So we have just seen an example where adjoining one root of a polynomial gives us all of the roots.
I hope this helps!