Let
$$M(d) = \frac{(a_1+d)(a_2+d)\dots(a_n+d)}{a_1 a_2 \dots a_n}.$$
The key to the proof is to notice that the assumption that $M(d)$ is an integer for every non-negative integer $d$ implies indeed that $M(d)$ is an integer for all integer $d$: if $d < 0$, consider $d' = |d| a_1 \dots a_n + d$; then, $d' \ge |d| + d \ge 0$ and so, $M(d')$ is integer. Then, $(a_1+d')(a_2+d')\dots(a_n+d') \equiv 0 \pmod{a_1 \dots a_n}$; since $d' \equiv d \pmod{a_1 \dots a_n}$, we deduce that $(a_1+d)(a_2+d)\dots(a_n+d) \equiv 0 \pmod{a_1 \dots a_n}$ and, therefore, $M(d)$ is an integer. We will now show that $a_k = k$ for every $1 \le k \le n$. We proceed inductively. Assuming that $a_i = i$ for every $i < k$, for $k \ge 1$, we will show that $a_k = k$. Consider $M(-k)$. By the induction hypothesis, we have that
$$M(-k) = \frac{(-1)^{k-1}(k-1)!(a_k-k)(a_{k+1}-k)\dots(a_n-k)}{(k-1)! a_k a_{k+1} \dots a_n} = \frac{(-1)^{k-1}(a_k-k)(a_{k+1}-k)\dots(a_n-k)}{a_k a_{k+1} \dots a_n}.$$
If $a_k > k$, then $0 < a_j - k < a_j$ for every $k \le j \le n$, and so, we have that
$$0 < (a_k - k)(a_{k+1} - k)\dots(a_n - k) < a_k a_{k+1} \dots a_n,$$
which implies that $0 < |M(-k)| < 1$. This contradicts the fact that $M(-k)$ is an integer. Therefore, $a_k = k$, which completes the induction.

We conclude that $a_k = k$ for every $1 \le k \le n$. To finish the proof, note that the condition in the statement holds for these values, since for every integer $d \ge 0$, we have $M(d) = \binom{n+d}{n}$, which is integer.