I have four rules here that are true and I wanted to make sure I have a general intuition as of why. These aren't meant to be rigorous proofs, but rather simple explanations.
Suppose $R$ is a regular expression. Then,
$R\cup \emptyset = R$. This is because $\emptyset$ is empty. This is straightforward.
$R\cdot\emptyset = \emptyset$. This is because concatenation is defined as $A\cdot B = \{xy | x\in A \texttt{ and } y\in B\}$. Nothing is in $\emptyset$, so our second requirement is not fulfilled, and never will be, so we get $\emptyset$.
$R \cdot\epsilon = R$. This is because $\epsilon$ doesn't change the regular regular expression in any way. (My explanation is lacking, and is not that convincing).
$R\cup \epsilon \neq R$. Suppose some machine $M$ recognizes RE $R$. This machine does not accept $\epsilon$. Then clearly this machine will not accept $R\cup\epsilon$, so $R\cup \epsilon \neq R$.
Do these generally sound right?
First of all, you make a confusion between regular expression and regular languages. That said, the formulas $R \cup \emptyset = R$ $R\emptyset = \emptyset$ and $R\varepsilon = R$ hold for every language $R$ (regular or not), but the formula $R \cup \varepsilon \neq R$ does not hold in general: just take $R = \varepsilon$ for a counterexample.