(Example 3 Given that the set $M$ consists of all functions $f(x)$ that satisfy the following property: there exists an $x_{0}$ in the domain such that $f\left(x_{0}+1\right)=f\left(x_{0}\right)+f(1)$.
(1) Does the function $f(x)=\frac{1}{x}$ belong to the set $M$? Explain your reasoning;
(2) Suppose the function $g(x)=\lg \frac{a}{x^{2}+1} \in M$, find the range of the real number $a$;
(3) Suppose the graph of the function $y=2^{x}$ intersects with the graph of the function $y=-x$, prove: the function $h(x)=$ $2^{x}+x^{2} \in M$.
(1) If $f(x)=\frac{1}{x} \in M$, then $\frac{1}{x_{0}+1}=\frac{1}{x_{0}}+1$, which means $x_{0}^{2}+x_{0}+1=0$. However, this equation has no real solutions. Therefore, $f(x)=\frac{1}{x} \notin M$.
(2) According to the problem, we know that $\lg \frac{a}{\left(x_{0}+1\right)^{2}+1}=\lg \frac{a}{x_{0}^{2}+1}+\lg \frac{a}{2}$ has real solutions, which means $\frac{a}{\left(x_{0}+1\right)^{2}+1}=\frac{a}{x_{0}^{2}+1} \cdot \frac{a}{2}$, or $(a-2) x_{0}^{2}+2 a x_{0}+2 a-2=0$ has real solutions.
When $a=2$, it clearly has a real solution $x_{0}=-\frac{1}{2}$;
When $a \neq 2$, we should have $\Delta=4 a^{2}-4(a-2)(2 a-2) \geqslant 0$, which gives $a \in[3-\sqrt{5}, 2) \cup(2, 3+\sqrt{5}]$.
In summary, $a \in[3-\sqrt{5}, 3+\sqrt{5}]$.
(3) The graph of the function $y=2^{x}$ intersects with the graph of the function $y=-x$, i.e., $\left\{\begin{array}{l}y=2^{x}, \\ y=-x,\end{array}\right.$, which means $2^{x}+x=0$ has a solution. Let the solution be $t$, and take $x_{0}=t+1$. Then
$$
h\left(x_{0}+1\right)=h\left(x_{0}\right)+h(1) \Leftrightarrow 2^{x_{0}+1}+\left(x_{0}+1\right)^{2}=\left(2^{x_{0}}+x_{0}^{2}\right)+3
$$
$$
\begin{array}{l}
\Leftrightarrow 2^{x_{0}}+2\left(x_{0}-1\right)=0 \\
\Leftrightarrow 2^{x_{0}-1}+\left(x_{0}-1\right)=0,
\end{array}
$$

Since $2^{t}+t=0$, i.e., $2^{x_{0}-1}+\left(x_{0}-1\right)=0$ holds.
Therefore, $h(x)=2^{x}+x^{2} \in M$.