Part 1: All polynomials with $\operatorname{deg} P=1$ satisfy the given property.
Suppose $P(x)=c x+d$, and assume without loss of generality that $c>d \geq 0$. Denote $s_{i}=a_{1}+a_{2}+ \cdots+a_{i} (\bmod c)$. It suffices to show that there exist indices $i$ and $j$ such that $j-i \geq 2$ and $s_{j}-s_{i} \equiv d \pmod{c}$.

Consider $c+1$ indices $e_{1}, e_{2}, \ldots, e_{c+1}>1$ such that $a_{e_{l}} \equiv d \pmod{c}$. By the pigeonhole principle, among the $n+1$ pairs $\left(s_{e_{1}-1}, s_{e_{1}}\right),\left(s_{e_{2}-1}, s_{e_{2}}\right), \ldots,\left(s_{n+1-1}, s_{n+1}\right)$, some two are equal, say $\left(s_{m-1}, s_{m}\right)$ and $\left(s_{n-1}, s_{n}\right)$. We can then take $i=m-1$ and $j=n$.

Part 2: All polynomials with $\operatorname{deg} P \neq 1$ do not satisfy the given property.
Lemma: If $\operatorname{deg} P \neq 1$, then for any positive integers $A, B$, and $C$, there exists an integer $y$ with $|y|>C$ such that no value in the range of $P$ falls within the interval $[y-A, y+B]$.

Proof of Lemma: The claim is immediate when $P$ is constant or when $\operatorname{deg} P$ is even since $P$ is bounded from below. Let $P(x)=a_{n} x^{n}+\cdots+a_{1} x+a_{0}$ be of odd degree greater than 1, and assume without loss of generality that $a_{n}>0$. Since $P(x+1)-P(x)=a_{n} n x^{n-1}+\ldots$, and $n-1>0$, the gap between $P(x)$ and $P(x+1)$ grows arbitrarily for large $x$. The claim follows. $\square$

Suppose $\operatorname{deg} P \neq 1$. We will inductively construct a sequence $\{a_{i}\}$ such that for any indices $i<j$ and any integer $k$ it holds that $a_{i}+a_{i+1}+\cdots+a_{j} \neq P(k)$. Suppose that we have constructed the sequence up to $a_{i}$, and $m$ is an integer with smallest magnitude yet to appear in the sequence. We will add two more terms to the sequence. Take $a_{i+2}=m$. Consider all the new sums of at least two consecutive terms; each of them contains $a_{i+1}$. Hence all such sums are in the interval $[a_{i+1}-A, a_{i+1}+B]$ for fixed constants $A, B$. The lemma allows us to choose $a_{i+1}$ so that all such sums avoid the range of $P$.

Alternate Solution for Part 1: Again, suppose $P(x)=c x+d$, and assume without loss of generality that $c>d \geq 0$. Let $S_{i}=\{a_{j}+a_{j+1}+\cdots+a_{i} (\bmod c) \mid j=1,2, \ldots, i\}$. Then $S_{i+1}=\{s_{i}+a_{i+1} (\bmod c) \mid s_{i} \in S_{i}\} \cup \{a_{i+1} (\bmod c)\}$. Hence $|S_{i+1}|=|S_{i}|$ or $|S_{i+1}|=|S_{i}|+1$, with the former occurring exactly when $0 \in S_{i}$. Since $|S_{i}| \leq c$, the latter can only occur finitely many times, so there exists $I$ such that $0 \in S_{i}$ for all $i \geq I$. Let $t>I$ be an index with $a_{t} \equiv d \pmod{c}$. Then we can find a sum of at least two consecutive terms ending at $a_{t}$ and congruent to $d \pmod{c}$.

Alternate Construction when $P(x)$ is constant or of even degree
If $P(x)$ is of even degree, then $P$ is bounded from below or from above. In case $P$ is constant or bounded from above, then there exists a positive integer $c$ such that $P(x)<c$. Let $\{a_{i}\}$ be the sequence
$$0,1,-1,2,3,-2,4,5,-3, \cdots$$
which is given by $a_{3 n+1}=2 n$, $a_{3 n+2}=2 n+1$, $a_{3 n+3}=-(n+1)$ for all $n \geq 0$. Notice that for any $i<j$ we have $a_{i}+\cdots+a_{j} \geq 0$. Then for the sequence $\{b_{n}\}$ defined by $b_{n}=a_{n}+c$, clearly $b_{i}+\cdots+b_{j} \geq (a_{i}+\cdots+a_{j})+2c>c$ which is outside the range of $P(x)$.

Now if $P$ is bounded from below, there exists a positive integer $c$ such that $P(x)>-c$. In this case, take $b_{n}=-a_{n}-c$. Then for all $i<j$ we have $b_{i}+\cdots+b_{j} \leq -(a_{i}+\cdots+a_{j})-2c<-c$ which is again outside the range of $P(x)$.