No, (1) and (2) are both false.

Take $n=1$ and $\Omega = (0,1)$ the unit interval.

**Lemma**. For $0 \le \delta \le \frac{1}{2 \sqrt{2}}$ we have $\sin(\pi x) \ge \delta \sin(2 \pi x)$ for all $x \in [0,1]$.

*Proof*. Set $F(x) = \sin(\pi x) - \delta \sin(2 \pi x)$. Note $F(0) = 0$. For $0 < x < \frac{1}{4}$ we have
$$\pi \cos(\pi x) \ge \pi \cos(\frac{\pi}{4}) = \frac{\pi}{\sqrt{2}} \ge 2 \pi \delta \ge 2 \pi \delta \cos(2 \pi x)$$
proving that $F'(x) \ge 0$. So $F$ is increasing on $(0,\frac{1}{4})$ and hence $F \ge 0$ on $(0,\frac{1}{4})$. If $\frac{1}{4} \le x \le \frac{1}{2}$ we have
$$\sin(\pi x) \ge \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} > \delta \ge \delta \sin(2 \pi x)$$
and so $F(x) \ge 0$ on $[\frac{1}{4}, \frac{1}{2}]$. Finally for $\frac{1}{2} \le x \le 1$ we have $$\sin(\pi x) \ge 0 \ge \delta \sin(2 \pi x)$$
and so $F(x) \ge 0$ on $[\frac{1}{2}, 1]$. QED.

Now set $f(x) = \sin(\pi x) - \sin(2 \pi x)$. Clearly $f(x) < 0$ for sufficiently small $x > 0$. Since $\sin(n \pi x)$ is a Dirichlet eigenfunction of $\Delta$ with eigenvalue $-(n \pi)^2$, we have $$(S(t) f)(x) = e^{-\pi^2 t} \sin(\pi x) - e^{-4 \pi^2 t} \sin (2 \pi x) = e^{-\pi^2 t}\left(\sin(\pi x) - e^{-3 \pi^2 t} \sin(2 \pi x)\right)$$
so if we choose $t$ large enough that $e^{-3 \pi^2 t} \le \frac{1}{2 \sqrt{2}}$ we have $S(t) f \ge 0$ by our lemma, disproving (1).

For (2), set $f(x) = \sin(2 \pi x)$ and $$g(x) = \begin{cases} 1, & 0 \le x < \frac{1}{2} \\ 0, & \frac{1}{2} \le x \le 1. \end{cases}$$ Clearly $fg \ge 0$. By the strong maximum principle, $S(t) g > 0$ everywhere for all $t > 0$, but $(S(t) f)(x) = e^{-4 \pi^2 t} \sin(2 \pi x) < 0$ for $\frac{1}{2} < x < 1$. So for any $t > 0$ we have $(S(t) f)(S(t) g) < 0$ on $(\frac{1}{2}, 1)$.