Uniqueness of determinant
If the determinant $\delta$ exists, you can prove that
$\delta(E)=c$ if $E$ is the elementary matrix corresponding to multiplication of a row by $c$;
$\delta(E)=1$ if $E$ is the elementary matrix corresponding to summing a row to another multiplied by a constant;
$\delta(E)=-1$ if $E$ is the elementary matrix corresponding to switching two rows;
$\delta(A)=0$ if $A$ is not invertible;
$\delta(AB)=\delta(A)\delta(B)$.
If two determinant functions $\delta$ and $\delta'$ exist, then they are both zero on the noninvertible matrices, but, writing an invertible $A$ as $$ A=E_1E_2\dots E_k $$ a product of elementary matrices, we conclude that $$ \delta(A)=\delta(E_1)\delta(E_2)\dots\delta(E_k)=\delta'(A) $$
Note the initial if: we're assuming the existence, not proving it. Also, note that the decomposition into a product of elementary matrices doesn't depend on the determinant; so just use the same decomposition for computing $\delta(A)$ and $\delta'(A)$.
The fact that the decomposition as product of elementary matrices is not unique is indeed a problem, but with respect to the existence of the determinant. In principle you could find two decompositions that produce different values when computed with rules 1–5, but this would simply prove that the determinant doesn't exist.