It is a sketch of the proof in the case of $p\ne q$ (a complete
proof on these lines is rather lengthy). Something similar can be
done in more general case, possibly with some exceptions.

Our plan is the following: We assume that $p,q\in[1,\infty)$,
$p\ne q$ and $W=E_1\oplus_p E_2=F_1\oplus_q F_2$ (isometrically),
where $E_1,E_2,F_1,F_2$ are all isometric to a Banach space $E$,
and get a contradiction.


By $S(X)$ we denote the unit sphere of a Banach space $X$, if $Z$
and $Y$ are subspaces of $X$, we set $\delta(Y,Z)=\inf\{||y-z||:~
y\in S(Y), z\in Z\}$ and call it an *inclination* of $Y$ to
$Z$.

If we can find $i\ne j\in\{1,2\}$ such that $F_i$ has zero
inclination to $E_1$ and $F_j$ has zero inclination to $E_2$, we
get a contradiction by observing that it means that
two-dimensional $\ell_p$-sphere approximates $\ell_q$-sphere with
an arbitrary precision, which is not true ($p$ and $q$ are fixed).

It remains to consider the case where both $F_1$ and $F_2$ have
nonzero inclination to $E_1$ (or to $E_2$, the cases are
similar).


By $P_1$ and $P_2$ we denote projections on $W$ corresponding to
the decomposition $E_1\oplus_p E_2$. Nonzero inclination to $E_1$
implies that the restriction of $P_2$ to both $F_1$ and $F_2$ are
isomorphic embeddings. If there are nonzero points  $y_1$ and
$y_2$ in $F_1$ and $F_2$ which have the same image in $E_2,$ we
get a contradiction by considering the space spanned by $y_1$ and
$y_2$: on one hand its unit sphere is $\ell_p$-sphere, and on the
other hand - $\ell_q$-sphere.

In a similar way (but using approximations) we get a contradiction
in the case where the inclination of $P_2(F_1)$ to $P_2(F_2)$ is
zero. Finally, if the inclination of $P_2(F_1)$ to $P_2(F_2)$ is
nonzero, we get a contradiction because in this case $P_2$ would
be an isomorphic embedding of $W$ into $E_2$, which is obviously
false.