= Solution
Throughout, the braces denote a multiset, so repeated splitting degrees retain their multiplicities. If the multisets agree after a common shift, permuting summands gives $E_a\cong E_b\otimes\mathcal O(c)$. Tensoring all the fibres of a <vector bundle> by the same <line bundle> leaves their one-dimensional quotient spaces unchanged. Hence
$$
\mathbb P(E_a)\cong\mathbb P(E_b).
$$
This proves one implication in the <isomorphism classification of abstract rational scrolls>.
For the converse first take $n\ge3$. Every divisor class on $F_a$ has the form $xL+yM$. If its square is zero, restriction to a fibre $\mathbb P^{n-1}$ gives
$$
y^2h^2=0,
$$
where $h$ is the hyperplane class and $h^2\ne0$ because $n-1\ge2$. Thus $y=0$. It follows that $L$ is intrinsically characterized as the primitive nonzero nef class whose square is zero. An <isomorphism> $u:F_a\to F_b$ therefore carries $L_b$ to $L_a$; the sign is positive because both classes are nef and effective. The complete pencil $|L|$ determines the ruling, so
$$
\pi_b\circ u=\sigma\circ\pi_a
$$
for an <automorphism> $\sigma$ of $\mathbb P^1$.
On a fibre, the <isomorphism> preserves the positive generator $\mathcal O_{\mathbb P^{n-1}}(1)$ of its <Picard group>. Consequently
$$
u^*M_b=M_a+cL_a
$$
for some integer $c$. Push forward along the ruling, using $\pi_*\mathcal O(M)=E$ and the <projection formula for sheaves>:
$$
\sigma^*E_b\cong E_a\otimes\mathcal O_{\mathbb P^1}(c).
$$
Every <automorphism> of $\mathbb P^1$ preserves each $\mathcal O(d)$. The uniqueness of the splitting degrees in the <Birkhoff–Grothendieck theorem> now implies equality of the two multisets up to a shift. Replacing this $c$ by $-c$ gives the sign convention in the final formula. One can see that uniqueness numerically: for a split bundle,
$$
h^0(E(t))=\sum_i\max(a_i+t+1,0).
$$
The first difference counts the degrees with $a_i+t\ge0$, and the second difference counts the multiplicity of $a_i=-t$. Hence these dimensions determine every splitting degree, with multiplicity.
For $n=2$, subtract the smaller degree and write the normalized bundle as $\mathcal O\oplus\mathcal O(a)$, $a\ge0$. Its surface is the <Hirzebruch surface> $\mathbb F_a$. Let $B$ be its negative section and $L$ its fibre, so $B^2=-a$, $BL=1$, $L^2=0$. The <nef cone> is generated by $L$ and $B+aL$: a <nef divisor> $xL+yB$ must have $y\ge0$ from intersection with a fibre and $x-ay\ge0$ from intersection with $B$, and these two inequalities also express it as a nonnegative combination of those generators. If $a>0$, a class $uL+v(B+aL)$ with $u,v\ge0$ has square
$$
2uv+av^2,
$$
which vanishes only when $v=0$. Thus the ruling is again intrinsic, and the same bundle argument applies. If $a=0$, the surface is $\mathbb P^1\times\mathbb P^1$ and has two nef square-zero rays, its two rulings. This distinguishes it from every $a>0$ surface. <Isomorphisms> may exchange its rulings, but both normalized splitting degrees are zero, so the stated classification is unchanged.
For $n=1$, every <scroll> is $\mathbb P^1$, and any two single splitting degrees differ by a common shift. Thus all ranks are covered:
$$
\boxed{\mathbb F(a_1,\ldots,a_n)\cong\mathbb F(b_1,\ldots,b_n)
\iff
\{a_1,\ldots,a_n\}_{\mathrm{mult}}
=\{b_1+c,\ldots,b_n+c\}_{\mathrm{mult}}
\text{ for some }c\in\mathbb Z.}
$$
Reading the braces as ordinary sets would lose information and make the assertion false when degrees repeat. For example, $(0,0,1,2)$ and $(0,1,1,2)$ have the same underlying set but not the same splitting multiset; their minima are both zero, so no nonzero common shift can reconcile them.
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