The first step is the following. Assume $E'$ and $E''$ are two objects with filtrations as in axiom (d). Then if $\phi_n(E') > \phi_1(E'')$ then $Hom(E',E'') = 0$. This follows from (c) by induction.
The second step. Assume $E$ and $F$ are two objects, $\phi \in \mathbb{R}$, and $E' \to E \to E''$, $F' \to F \to F''$ are two triangles, such that $E'$, $F'$, $E''$, $F''$ have filtrations as in axiom (d) with
$$
\phi_n(E') > \phi \ge \phi_1(E''),\qquad
\phi_m(F') > \phi \ge \phi_1(F'').
$$
Then for any morphism $f:E \to F$ there are unique morphisms $f':E' \to F'$ and $f'':E'' \to F''$ such that $(f',f,f'')$ is a morphism of triangles. Existence follows from $Hom(E',F'') = 0$ and axioms of triangulated category. Uniqueness follows from long exact sequence of $Ext$'s and $Hom(E',F''[-1]) = 0$ (note that the phases of $F''[-1]$ are strictly smaller than those of $F''$ by (b)).
Note, that from the uniqueness above it follows that if $f$ is an isomorphism, then so are $f'$ and $f''$. Indeed, if $g = f^{-1}$, then there are unique $g'$ and $g''$, and since $(f'\circ g',f\circ g,f'',circ g'')$ and $(1_{F'},1_F,1_{F''})$ are morphisms of triangles with the same middle map, uniqueness implies that $f'\circ g' = 1_{F'}$ and $f'' \circ g'' = 1_{F''}$.
Now one can prove a uniqueness of the filtration. Assume there are two.
Choose some $\phi$ and split the filtrations at $\phi$. Take $f = 1$, and extend it to $(f',f,f'')$. By the argument above $f'$ and $f''$ are isomorphisms. Continue by induction.
Finally, you cannot artificially add $A_i = 0$, as $P(\phi)$ never contains $0$ by axiom (a), as $m(E)$ is assumed to be strictly positive!
