W.l.o.g. no two of the $P_i$ are equal. Then there are only finitely many $x$ with $P_{i_1}(x)=P_{i_2}(x)$ for some $i_1 \ne i_2$. In particular, for all but finitely many $y$, the size of $S_y$ is just the sum of the sizes of the $P_i^{-1}(y)$. Hence, if $n_1$ of the $P_i$ have odd degree and $n_2$ of the $P_2$ have even degree, we have that $\vert S_y\vert=n_1+2n_2$ as $y \to \infty$ and $\vert S_y\vert=n_1$ as $y \to -\infty$. Hence $n_2=0$ i.e. all $P_i$ have odd degree. Moreover, $\vert S_y\vert=n$ for all $y$. Since now all $P_i$ are surjective, for all but finitely many $y$, each $P_i$ takes the value $y$ exactly once. But then this must actually be the case for all $y$, i.e. all $P_i$ are strictly increasing. But then for $\vert S_y\vert=n$ to actually hold for all $y$, we can never have $P_{i_1}(x)=P_{i_2}(x)$ for $i_1 \ne i_2$ since otherwise the size of $S_y$ decreases. Hence any difference $P_{i_1}-P_{i_2}$ must not have a real root, hence must be an even degree polynomial. So the leading terms must cancel i.e. any two of the $P_i$ must have the same degree as desired.

Claim 1: all polynomials have odd degree. Proof: Let there be some even degree polynomials.Let the largest extremal value of all $P_i$ be $T$, and the smallest be $T'$. Now take $y_1=T+1$ and $y_2=T'-1$ to find that $|S_{y_1}|>|S_{y_2}|$. From now on we define $L$ to be the largest value such that $|S_L|\neq n$ Claim 2: all polynomials are injective. Proof: Let $u$ be the largest value where there is two values $v_1,v_2$ that gives $P_i(v_1)=P_i(v_2)=u$ among all $i$, then take $y_1=u$ and $y_2=L+1$ to find out $|S_{y_1}|>|S_{y_2}|=n$. If this fails due to an intersection of two polynomials, then replace $u$ with $u-\epsilon$. Claim 3: The polynomials does not intersect. Proof: since all polynomials are injective, there's an unique $u$ such that $P_i(u)=v$ for each $v$. If there's a intersection at $y=w$ such that $P_i(w)=P_j(w)$, then take $y_1=w$ and $y_2=L+1$ to find out $|S_{y_1}|<|S_{y_2}|=n$. By claim 3 the polynomials does not intersect, so they have the same degree.