Proving properties of a system containing only two objects (even and odd numbers). Working on the book: Lange, Serge. "Basic Mathematics" (p. 63, exercise 1).

  
*
  
*Let E be an abbreviation for even, and let I be an abbreviation for odd. We know that:
  
  
  $$E+E=E\\
E + I = I + E = I\\
I + I = E\\
EE = E\\
II = I\\
IE = EI = E$$
a) Show that addition for E and I is associative and commutative. Show that E plays the role of a zero element for addition. What is the additive inverse of E? What is the additive inverse of I?
b) Show that multiplication for E and I is commutative and associative. Which of E or I behaves like 1? Which behaves like 0 for multiplication? Show that multiplication is distributive with respect to addition.

How can I approach this exercise ? Do I need to do an exhaustive proof or I can assume arbitrary elements belong to this system and proceed from there ?
EDIT: Based on the excellent suggestion received, I came up with this proof. Is this approach correct ?
a)
Addition for $E$ and $I$  is associative.
$
\begin{align*}
(E+E)+E=E+E&=\color{blue}E=E+E=E+(E+E)\\
(E+E)+I=E+I&=\color{blue}I=E+I=E+(E+I)\\
(E+I)+E=I+E&=\color{blue}I=E+I=E+(I+E)\\
(E+I)+I=I+I&=\color{blue}E=E+E=E+(I+I)\\
(I+E)+E=I+E&=\color{blue}I=I+E=I+(E+E)\\
(I+E)+I=I+I&=\color{blue}E=I+I=I+(E+I)\\
(I+E)+I=I+I&=\color{blue}E=I+I=I+(E+I)\\
(I+I)+I=E+I&=\color{blue}I=I+E=I+(I+I)\\
\end{align*}
$
Addition for $E$ and $I$ is commutative.
$
\begin{align*}
E+E&=\color{blue}E=E+E\\
E+I&=\color{blue}I=I+E\\
I+E&=\color{blue}I=E+I\\
I+I&=\color{blue}E=I+I\\
\end{align*}
$
$E$ is the zero element for addition.
$
E+a=a+E=a\\
\begin{align*}
E+\color{blue}E&=\color{blue}E+E=a\\
E+\color{blue}I&=\color{blue}I+E=a\\
\end{align*}
$
$E$ is the additive inverse of $E$.
$
E+a=a+E=a\\
\begin{align*}
E+\color{blue}E=\color{blue}E+E=E
\end{align*}
$
b)
Multiplication for $E$ and $I$  is associative.
$
\begin{align*}
(EE)E=EE&=\color{blue}E=EE=E(EE)\\
(EE)I=EI&=\color{blue}E=EI=E(EI)\\
(EI)E=IE&=\color{blue}E=EI=E(IE)\\
(EI)I=II&=\color{blue}I=EE=E(II)\\
(IE)E=IE&=\color{blue}E=IE=I(EE)\\
(IE)I=II&=\color{blue}I=II=I(EI)\\
(IE)I=II&=\color{blue}I=II=I(EI)\\
(II)I=EI&=\color{blue}E=IE=I(II)\\
\end{align*}
$
Multiplication for $E$ and $I$ is commutative.
$
\begin{align*}
EE&=\color{blue}E=EE\\
EI&=\color{blue}E=IE\\
IE&=\color{blue}E=EI\\
II&=\color{blue}I=II\\
\end{align*}
$
$I$ is the 1 element for multiplication.
$
Ia=aI=a\\
\begin{align*}
I\color{blue}E&=\color{blue}EI=\color{blue}E\\
I\color{blue}I&=\color{blue}II=\color{blue}I\\
\end{align*}
$
$E$ is the zero element for multiplication.
$
Ea=aE=E\\
\begin{align*}
E\color{blue}E&=\color{blue}EE=\color{blue}E\\
E\color{blue}I&=\color{blue}IE=\color{blue}E\\
\end{align*}
$
Multiplication is distributive with respect to addition.
$
\begin{align*}
E(E+E)=EE&=\color{blue}E=E+E=EE+EE\\
E(E+I)=EI&=\color{blue}E=E+E=EE+EI\\
E(I+E)=EI&=\color{blue}E=E+E=EI+EE\\
E(I+I)=EI&=\color{blue}E=E+E=EI+EI\\
I(E+E)=IE&=\color{blue}E=E+E=IE+IE\\
I(E+I)=II&=\color{blue}I=E+I=IE+II\\
I(E+I)=II&=\color{blue}I=I+E=II+IE\\
I(I+I)=IE&=\color{blue}E=I+I=II+II\\
\end{align*}
$
 A: You only have 6 equations in the table.  Just do them all:

a) Show that addition for E and I is associative:

So show $(a+b) + c= a+(b+ c)$.
By considering the eight cases of $a = E,I; b=E,I; c=E,I$.

and commutative.

So show $a+b = b+a$.
By considering the four cases of $a = E,I; b=E; I$.
(Although there are some logical shortcuts you can take).

Show that E plays the role of a zero element for addition. 

So show $E+a = a+E = a$ by considering the two case $a=E;I$.

What is the additive inverse of E?

So show there is exactly one element, $a$, where $E+a = a+E = E$.  Do this by looking and the six line table you listed above.

What is the additive inverse of I?

ditto but for $a+I = I + a = E$.
A: I don't know what you mean by "assume arbitrary elements belong to this system." There are two, and only two, elements in the system: $E$ and $I$. You should do an exhaustive proof, but it won't be too exhausting: there aren't that many possible additions and multiplications of merely two elements. 
