Using De Morgans rule I can easily prove $E\cap F$ is an element of $\mathcal{A}$. This is done by applying the following definition and the three axioms.
Definition: Let $X$ be some set. An algebra of subsets of $X$, $\mathcal{A}$, is a collection of subsets of $X$ which satisfy the following three axioms;
$X \in \mathcal{A}$
If $E \in \mathcal{A}$ then $E^c \in \mathcal{A}$
If $E, F \in \mathcal{A}$ then $E \cup F \in \mathcal{A}$
However my question is can these axioms be used to show if $E, F\in \mathcal{A}$ then $E\setminus F \in \mathcal{A}$ ?
Suppose that $E$ and $F$ are both elements of $\mathcal{A}$
Since $E$ is an element of $\mathcal{A}$ and $\mathcal{A}$ is closed under complementation, it follows that $E^c$ is an element of $\mathcal{A}$.
Further, since $E^c$ and $F$ are both elements of $\mathcal{A}$ and $\mathcal{A}$ is closed under union, it follows that $(E^c\cup F)$ is an element of $\mathcal{A}$.
Finally, noting that $(E^c\cup F)$ is an element of $\mathcal{A}$ and that $\mathcal{A}$ is closed under complementation, it follows that $(E^c\cup F)^c$ is an element of $\mathcal{A}$.
Then noting that: $E\setminus F = E\cap F^c = (E^c\cup F)^c$, it follows that $E\setminus F$ is an element of $\mathcal{A}$ and therefore $\mathcal{A}$ is closed under set differences.