1. To prove $f(x) \geqslant x - x^2$, we consider the function $h(x) = \ln(x+1) - x + x^2$ for $x \geqslant 0$. We calculate the derivative of $h(x)$ as follows:
$$h'(x) = \frac{d}{dx}[\ln(x+1) - x + x^2] = \frac{1}{x+1} - 1 + 2x.$$
Simplifying the derivative, we get:
$$h'(x) = \frac{1}{x+1} + 2x - 1 = \frac{2x^2 + x}{x+1} \geqslant 0.$$
Since $h'(x) \geqslant 0$ for all $x \geqslant 0$, $h(x)$ is monotonically increasing on $[0, +\infty)$. Evaluating $h(x)$ at $x=0$, we find:
$$h(0) = \ln(1) - 0 + 0^2 = 0.$$
Therefore, since $h(x)$ is monotonically increasing and $h(0) = 0$, we have $h(x) \geqslant 0$ for all $x \geqslant 0$. This proves that $\ln(x+1) \geqslant x - x^2$.

2. For the inequality $f(x) + x \geqslant g(x)$ to hold, we rewrite it as $\ln(x+1) \geqslant \frac{ax}{1+x}$. Let $m(x) = \ln(x+1) - \frac{ax}{1+x}$. We want to ensure $m(x) \geqslant 0$ for all $x \geqslant 0$. Calculating the derivative of $m(x)$, we get:
$$m'(x) = \frac{1}{x+1} - \frac{a(1+x) - ax}{(1+x)^2} = \frac{x+1-a}{(1+x)^2}.$$
For $m'(x) > 0$, we need $x+1-a > 0$, which simplifies to $x > a-1$. If $a \leqslant 1$, then $m(x)$ is monotonically increasing on $[0, +\infty)$, and since $m(0) = 0$, $m(x) \geqslant 0$ always holds. If $a > 1$, then $m(x)$ is monotonically decreasing on $[0, a-1]$ and increasing on $(a-1, +\infty)$. In this case, $m(x)_{\min} = m(a-1) x - x^2$. Letting $x = \frac{1}{n}$, where $n \in \mathbb{N}^*$, we get:
$$\ln\left(\frac{n+1}{n}\right) > \frac{n-1}{n^2}.$$
This implies:
$$\ln(n+1) - \ln(n) > \frac{n-1}{n^2}.$$
Summing these inequalities for $n = 1$ to $n$, we obtain:
$$\ln(n+1) > \frac{1}{4} + \frac{2}{9} + \ldots + \frac{n-1}{n^2}.$$
Since $n^2 + 3n + 2 > n + 1$, it follows that $\ln(n^2 + 3n + 2) > \ln(n+1)$. Therefore, for all $n \in \mathbb{N}^*$, we have:
$$\ln(n^2 + 3n + 2) > \frac{1}{4} + \frac{2}{9} + \ldots + \frac{n-1}{n^2}.$$
Thus, the statement is proven, and the inequality holds for all $n \in \mathbb{N}^*$.