Note that if for some real numbers $x_1, x_2, \dots, x_m$ the following equality holds:
$$|x_1| + |x_2| + \dots + |x_m| = |x_1 + x_2 + \dots + x_m|,$$
then they are of the same sign.
Let
$$Q(x) = P(x)(x-1)^t (x+1)^s = b_n x^n + b_{n-1} x^{n-1} + \dots + b_0,$$
where $b_n = a_d > 0$. From the problem statement it follows that
$$|b_0| = |b_1| + |b_2| + \dots + |b_n|.$$
Lemma: If $t \ge 1$, then $b_1, b_2, \dots, b_n \ge 0$.
Proof:
$$\begin{align*} Q(1) = 0 &\implies b_0 + b_1 + \dots + b_n = 0 \\ &\implies |b_1 + b_2 + \dots + b_n| = |b_0| = |b_1| + |b_2| + \dots + |b_n|, \end{align*}$$
hence, $b_1, b_2, \dots, b_n$ are of the same sign. Since $b_n > 0$, then $b_1, b_2, \dots, b_{n-1} \ge 0$. We proved the lemma.
Assume that $t \ge 2$. According to the lemma,
$$b_1, b_2, \dots, b_{n-1} \ge 0 \implies b_1 + 2b_2 + \dots + nb_n > 0.$$
On the other hand, let $R(x) = \frac{Q(x)}{(x-1)^2}$. Then
$$Q'(x) = 2(x-1)R(x) + (x-1)^2 R'(x) \implies Q'(1) = 0 \implies b_1 + 2b_2 + \dots + nb_n = 0$$
— a contradiction. Thus, $t \le 1$.

Similarly, we can show that $s \le 1$. Let's consider three cases.

**I) $t = 1, s = 1$**
$$\begin{align*} Q(x) = P(x)(x^2 - 1) &\implies Q(1) = Q(-1) = 0 \\ &\implies b_0 + b_1 + \dots + b_n = b_0 - b_1 + b_2 + \dots + (-1)^n b_n = 0 \\ &\implies b_1 + b_2 + b_3 + \dots = 0. \end{align*}$$
According to the lemma,
$$b_1, b_2, \dots, b_n \ge 0 \implies b_1 = b_2 = b_3 = \dots = 0,$$
— a contradiction, since $b_1 = -a_1$ and by the problem statement $a_1 \ne 0$.

**II) $t = 1, s = 0$**
$$Q(x) = P(x)(x-1) = -a_0 + (a_0 - a_1)x + \dots + (a_{d-1} - a_d)x^d + a_d x^{d+1}.$$
By the lemma,
$$a_0 - a_1 = b_1 \ge 0, \dots, a_{d-1} - a_d = b_d \ge 0 \implies a_0 \ge a_1 \ge \dots \ge a_d.$$

Therefore, $P(x)$ is non-increasing.

**III) $t=0, s=1$**.
$$Q(-1) = 0 \implies b_0 - b_1 + \dots + (-1)^n b_n = 0 \implies \\ \implies |b_1 - b_2 + \dots + (-1)^n b_n| = |b_0| = |b_1| + |-b_2| + \dots + |(-1)^n b_n|.$$
Thus, $b_1, -b_2, \dots, (-1)^n b_n$ are of the same sign, and since $b_n > 0$, then $(-1)^{n-i}b_i \ge 0$ for each $1 \le i \le n$.
$$Q(x) = P(x)(x+1) = a_0 + (a_0 + a_1)x + \dots + (a_{d-1} + a_d)x^d + a_d x^{d+1} \implies \\ \implies (-1)^{d+1-i}(a_{i-1} + a_i) \ge 0 \text{ (for all } 1 \le i \le d) \implies \\ \implies a_d \le -a_{d-1} \le a_{d-2} \le \dots \le (-1)^d a_0.$$
It follows that $(-1)^d P(-x) = a_d x^d - a_{d-1} x^{d-1} + \dots + (-1)^d a_0$ is non-increasing.