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The answer is $\theta : p \mapsto p(c)$, for each choice of $c \in \mathbb{Z}$. Obviously these work, so we prove these are the only ones. In what follows, $x \in \mathbb{Z}[x]$ is the identity polynomial, and $c = \theta(x)$.

First solution (Merlijn Staps) Consider an integer $n \neq c$. Because $x - n \mid p(x) - p(n)$, we have
$$\theta(x-n) \mid \theta(p(x)-p(n)) \implies c-n \mid \theta(p(x))-p(n).$$
On the other hand, $c-n \mid p(c)-p(n)$. Combining the previous two gives $c-n \mid \theta(p(x))-p(c)$, and by letting $n$ large we conclude $\theta(p(x))-p(c) = 0$, so $\theta(p(x)) = p(c)$.

Second solution First, we settle the case $\deg p = 0$. In that case, from the second property, $\theta(m) = m + \theta(0)$ for every integer $m \in \mathbb{Z}$ (viewed as a constant polynomial). Thus $m + \theta(0) \mid 2m + \theta(0)$, hence $m + \theta(0) \mid -\theta(0)$, so $\theta(0) = 0$ by taking $m$ large. Thus $\theta(m) = m$ for $m \in \mathbb{Z}$.
Next, we address the case of $\deg p = 1$. We know $\theta(x+b) = c+b$ for $b \in \mathbb{Z}$. Now for each particular $a \in \mathbb{Z}$, we have
$$c+k \mid \theta(x+k) \mid \theta(ax+ak) = \theta(ax) + ak \implies c+k \mid \theta(ax) - ac.$$
for any $k \neq -c$. Since this is true for large enough $k$, we conclude $\theta(ax) = ac$. Thus $\theta(ax+b) = ac+b$.
We now proceed by induction on $\deg p$. Fix a polynomial $p$ and assume it's true for all $p$ of smaller degree. Choose a large integer $n$ (to be determined later) for which $p(n) \neq p(c)$. We then have
$$\frac{p(c)-p(n)}{c-n} = \theta\left(\frac{p-p(n)}{x-n}\right) \mid \theta(p-p(n)) = \theta(p)-p(n).$$
Subtracting off $c-n$ times the left-hand side gives
$$\frac{p(c)-p(n)}{c-n} \mid \theta(p)-p(c).$$
The left-hand side can be made arbitrarily large by letting $n \to \infty$, since $\deg p \geq 2$. Thus $\theta(p) = p(c)$, concluding the proof.