Can't we factor out a constant in the cross product? I have the vectors $A=a\hat e_x$ and $B=a\hat e_y$, so 
$$
A\times B =
\begin{vmatrix}
\hat e_x &\hat e_y & \hat e_z \\
a & 0 & 0\\
0 & a &0
\end{vmatrix}=\hat za^2
$$
Q1: But why is the following wrong
\begin{align}
A\times B &=a\hat e_x\times a\hat e_y\\
&=(a\hat e_x)\times (a\hat e_y) \tag 1\\ 
&=(a\hat e_x)\times (\hat e_ya) \tag 2\\
&=a(\hat e_x)\times (\hat e_y)a \tag 3\\
&=a((\hat e_x)\times (\hat e_y)) \tag 4\\
&=a(\hat e_x\times \hat e_y) \tag 5\\
&=a\hat e_z \quad ?
\end{align}
Q2: Also, why is this wrong
\begin{align}
A\times B &=
\begin{vmatrix}
\hat e_x &\hat e_y & \hat e_z \\
a & 0 & 0\\
0 & a &0
\end{vmatrix} \\
&=
a
\begin{vmatrix}
\hat e_x &\hat e_y & \hat e_z \\
1 & 0 & 0\\ \tag 6
0 & 1 &0
\end{vmatrix}\\
&=a\hat z \quad ?
\end{align}
 A: 
$$=a(\hat e_x)\times (\hat e_y)a \tag 3$$
$$=a((\hat e_x)\times (\hat e_y)) \tag 4$$

What happens here...?
You may be mixing it up with the distributive property of the cross product over addition.

Q2: Also, why is this wrong
\begin{align}
A\times B &=
\begin{vmatrix}
\hat e_x &\hat e_y & \hat e_z \\
a & 0 & 0\\
0 & a &0
\end{vmatrix} \\
&=
a
\begin{vmatrix}
\hat e_x &\hat e_y & \hat e_z \\
1 & 0 & 0\\ \tag 6
0 & 1 &0
\end{vmatrix}\\
&=a\hat z \quad ?
\end{align}

For starters, the determinant notation for computing the cross product is actually just a mnemonic. But if you want to use properties of determinants, you should note that the determinant is linear in its columns and rows, so for example:
$$\begin{vmatrix}
a & b & c \\
\color{purple}{k}d& \color{purple}{k}e & \color{purple}{k}f \\
g & h & i
\end{vmatrix}=\color{purple}{k}\begin{vmatrix}
a & b & c \\
d& e & f \\
g & h & i
\end{vmatrix}$$
This means that if you want to factor out the $a$, you do that for the second and the third row:
$$\begin{vmatrix}
\hat e_x &\hat e_y & \hat e_z \\
\color{blue}{a}& 0 & 0\\
0 & \color{red}{a} &0
\end{vmatrix}=\color{blue}{a}\color{red}{a}\begin{vmatrix}
\hat e_x &\hat e_y & \hat e_z \\
1 & 0 & 0\\
0 & 1 &0
\end{vmatrix}=a^2\begin{vmatrix}
\hat e_x &\hat e_y & \hat e_z \\
1 & 0 & 0\\
0 & 1 &0
\end{vmatrix}$$
A: We have that by cross product definition 
$$A\times B =a\hat e_x\times a\hat e_y=a^2( e_x\times \hat e_y)$$
and by the properties of the determinant
$$A\times B =
\begin{vmatrix}
\hat e_x &\hat e_y & \hat e_z \\
a & 0 & 0\\
0 & a &0
\end{vmatrix} =
a^2
\begin{vmatrix}
\hat e_x &\hat e_y & \hat e_z \\
1 & 0 & 0\\ \tag 6
0 & 1 &0
\end{vmatrix}$$
A: Your first argument is wrong because you lost an $a$ between lines (3) and (4).
Your second argument is wrong because line (6) should have $a^2$, not $a$. You can confirm this just by evaluating the determinant, as you did at the beginning of your post. You have somehow misunderstood the rules for removing common factors from determinants.
