I've once asked a similar question only about groups, but I am interested whether the logic is still sound:
$(1)$Let $S$ be a generating set of a ring $R$, and denote $\kappa=\vert S\vert$. Then $\vert R\vert \leq \kappa$, and if $\kappa<\aleph_0$ then $\vert R \vert \leq \aleph_0$.
This is because any element is written as a finite sum of finite products of elements in $S$. Thus if we denote:
$R_{n,m}:=\Big\{ \underset{i=1}{\overset{n}{\sum}} \underset{j=1}{\overset{m_i}{\prod}} a_{i,j} \Big\vert m_i\leq m, \; \{ a_{i,j} \}\subseteq S \Big\}$
We can see that $R=\underset{n,m=1}{\overset{\infty}{\cup}}R_{n,m}$. Hence:
$\vert R\vert\leq \underset{n=1}{\overset{\infty}{\sum}}\underset{m=1}{\overset{\infty}{\sum}}\vert R_{n,m}\vert$. Where if $\kappa\geq \aleph_0$ then $\vert R_{n,m}\vert=\kappa$, and if $\kappa<\aleph_0$ then $\vert R_{n,m\vert } \leq \aleph_0$.
$(2)$The natural follow up question (using similar logic) is for a Module $M$ over a ring $R$, if $S$ generates the module $M$ such that $\vert S\vert= \kappa\geq \aleph_0$ is it true that:
$\vert M \vert=\kappa$ if $\kappa \geq \vert R\vert$, and $\vert M\vert=\vert R\vert$ if $\vert R\vert>\kappa$.
Let's work under the assumption that both $|R|$ and $|S|$ are infinite. You can work out the other cases.
There is a surjective map from the set $\Sigma_{<\omega}(R\times S)$ of finite sequences from the set $R\times S$ onto the set $M$. By standard cardinality arguments, $$ |\Sigma_{<\omega}(R\times S)|=|R\times S| $$ so we can conclude that $$ |S|\le|M|\le|R\times S| $$
It is known that $$ |R\times S|=\begin{cases} |R| & \text{if $|R|>|S|$} \\[4px] |S| & \text{if $|S|\ge|R|$} \end{cases} $$
Therefore, if $|R|\le|S|$ you can argue that $|M|=|S|$.
If $|S|<|R|$, one can have $|M|<|R|$. Let $R=J_p$ be the ring of $p$-adic integers; then $R/pR$ is finite (isomorphic to $\mathbb{Z}/p\mathbb{Z}$) and we can consider the module $M=(R/pR)^{(\aleph_0)}$ (countable direct sum of copies of $R/pR$). Then we can take $S=M$ and $|S|=|M|=\aleph_0$, but $|R|=2^{\aleph_0}$.
What's the obstruction? There is no injective homomorphism (no injective map, actually) $R\to M$, in this case.