Let's prove each part step-by-step.

(1) We are given that $a_n + S_n = n$. It follows that for $n=1$, we have $a_1 + S_1 = 1$, which simplifies to $2a_1 = 1$, hence $a_1 = \frac{1}{2}$. For $n + 1$, we have $a_{n+1} + S_{n+1} = n + 1$. Subtracting the nth equation from the n+1th, we get:
$$a_{n+1} + S_{n+1} - (a_n + S_n) = n+1 - n$$
$$a_{n+1} - a_n = 1$$
Since $S_{n+1} = S_n + a_{n+1}$, the left side can be written as $2a_{n+1} - (a_n + S_n) = 1$. This implies $2a_{n+1} - n = 1$, so $2(a_{n+1} - 1) = a_n - 1$. Therefore, the ratio between consecutive terms $c_{n+1}$ and $c_n$ is:
$$\frac{c_{n+1}}{c_n} = \frac{a_{n+1} - 1}{a_n - 1} = \frac{1}{2}$$
Given that $c_1 = a_1 - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$, we conclude that $\{c_n\}$ is a geometric sequence with first term $-\frac{1}{2}$ and common ratio $\frac{1}{2}$.

(2) Knowing that $\{c_n\}$ is a geometric sequence with first term $-\frac{1}{2}$ and common ratio $\frac{1}{2}$, we can write:
$$c_n = -\frac{1}{2^n}$$
And thus:
$$a_n = c_n + 1 = 1 - \frac{1}{2^n}$$
For $a_{n-1}$, it follows that:
$$a_{n-1} = 1 - \frac{1}{2^{n-1}}$$
For $n \geq 2$, we have:
$$b_n = a_n - a_{n-1} = \left(1 - \frac{1}{2^n}\right) - \left(1 - \frac{1}{2^{n-1}}\right) = \frac{1}{2^n}$$
For $n = 1$, we simply have $b_1 = a_1 = \frac{1}{2}$. Therefore, the general term for $\{b_n\}$ is:
$$b_n = \frac{1}{2^n}$$
For all $n \in \mathbb{N}^*$ (positive integers).

The general formula for the terms of the sequence $\{b_n\}$ is thus:
$$b_n = \boxed{\frac{1}{2^n}}$$
for all $n \in \mathbb{N}^*$.