I prefer to define a monotone operator on $\mathbb R^n$ to be a set-valued function $T:\mathbb R^n \to S$, where $S$ is the collection of all subsets of $\mathbb R^n$, such that
if $y_1 \in T(x_1)$ and $y_2 \in T(x_2)$ then
$$
\langle y_1 - y_2, x_1 - x_2 \rangle \geq 0.
$$
The graph of a set-valued function $T$ is $\{(x,y) \mid y \in T(x) \}$, which is a relation. Some authors define a monotone operator to be the graph itself (which is a relation), but to me it seems more clear to think of a monotone operator as being a set-valued function. (Reason: the subdifferential $\partial f$ is a convex function $f$ is supposed to be the prototypical example of a monotone operator, and in my mind $\partial f$ is a set-valued function, not a relation.)
Note that Bauschke and Combettes, one of the standard references for monotone operator theory, defines a monotone operator to be a function $A: \mathcal H \to 2^{\mathcal H}$ which satisfies the monotonicity property. (Here $\mathcal H$ is a Hilbert space.)
If $f$ is convex and differentiable on $\mathbb R^n$ then it can be shown that $\partial f(x) = \{\nabla f(x)\}$ for all $x \in \mathbb R^n$. So $\partial f(x)$ is a singleton for all $x$. It can also be shown that
$$
\langle \nabla f(x_1) - \nabla f(x_2), x_1 - x_2 \rangle \geq 0
$$
for all $x_1,x_2 \in \mathbb R^n$. However, if we are being very precise, then it would not be technically correct to say that $\nabla f$ is a monotone operator, because $\nabla f$ is not a set-valued function. It is certainly true, though, that $\partial f$ is a monotone operator. And most people will be a little sloppy and say that $\nabla f$ is a monotone operator, in this situation.
You also mentioned the function $g:\mathbb R^n \to \mathbb R^n$ defined by $g(x) = x$. As you mentioned,
$$
\langle g(x_1) - g(x_2), x_1 - x_2 \rangle \geq 0
$$
for all $x_1,x_2 \in \mathbb R^n$.
Again, if we are being very precise, it would not be technically correct to say that $g$ is a monotone operator, because $g$ is not a set-valued function. However, the operator defined by $\tilde g(x) = \{ g(x) \}$ is certainly a monotone operator. Most people will just be a little sloppy and state that $g$ itself is a monotone operator.
You also asked about the meaning of $(I + \lambda \nabla f)^{-1}$. It should really be $(I + \lambda \partial f)^{-1}$. Note that $I + \lambda \partial f$ is a set-valued function. And any set-valued function $T$ has an inverse $T^{-1}$ defined by
$$
T^{-1}(y) = \{x \mid y \in T(x) \}.
$$
So $T^{-1}$ is also a set valued function.
Thus, strictly speaking, $(I + \lambda \partial f)^{-1}$ is a set-valued function. But, there is an important fact: If the set-valued operator $T$ is maximal monotone, then $(I + \lambda T)^{-1}(y)$ is a singleton for all $y$. For this reason, we can view $(I + \lambda T)^{-1}$ as being a function that takes a vector as input and returns a single vector as output. This is a bit sloppy and not perfectly correct, but it's very common.
Moreover, if $f$ is a proper closed convex function, then $\partial f$ is a maximal monotone operator, so $(I + \lambda \partial f)^{-1}(y)$ is a singleton for all $y$, and so we can view $(I + \lambda \partial f)^{-1}$ as a function from $\mathbb R^n$ to $\mathbb R^n$. (It is not strictly correct, but it is harmless.)
