[This answer has been heavily edited in response to a long chain of comments on Eric Stucky's answer.]
I came up with these few theorems and I am curious whether or not my hypothesis are true. I don't really have a method for proving these, mind you. I'm more considering these as things I've noticed to be true. I just don't know whether they are always true.
Some of these are definitions to which I will append "def." to the end of the title. Think of these as a given. It's like how the integral is defined abstractly as area. I'm defining terms.
Jump Series Theorem def.
A function $f$ [consisting of floor] is said to have a jump series if there is a series $\Sigma=\sum_{n=1}^\infty a_nH_n$ of scaled step functions such that $f+\Sigma$ no longer has any discontinuity attributable to floor. The sequence of scaling factors $a_n$ is called the jump sequence for the function.
Definition of Implied Derivative def.
We define the evaluation of the implied derivative of a function $y$ at the point $x$ as the set of values $y^{\to}(x) = \{y | y = \lim_{h \to 0^+} \frac {y(x+h) - f(h)}{h} \lor y = \lim_{h \to 0^-} \frac {y(x+h) - f(h)}{h}\}$. It is in this sense that the implied derivative has the potential of being a multi-valued operator.
Jump Series Resolution Theorem
[The original wording is preserved; I have a formalism but it assumes that having a jump function is equivalent to being piecewise continuous, which I'm not sure is true.]
The integral of a floor-based function is floor treated as a constant minus the appropriate jump series of the integral.
Integration Continuity Theorem
The (ordinary) indefinite integral of any function $f$ having a jump series must be continuous (or fixably discontinuous) on the domain of $f$. However, this can vary for the implied indefinite integral.
Calculus Jump Theorem def.
[Again, the original wording is preserved; the meaning of this phrase has not been properly illuminated]
Any form of jumping within a function representable as a floor function can be considered as a portion of the constant of the implied integration for floor.
Jump Location Theorem
For any function $a(x)$ with a jump series, the function $\lfloor a(x)\rfloor$, has points of discontinuity precisely at those $x$ such that $a(x) - [a(x)] = 0$.
Composite Floor Function Jump Sequence Theorem One
If $a$ is a function such that $a(x) - \lfloor a(x) \rfloor = 0$ when $x - \lfloor x \rfloor = 0$, then the jump sequence of the composite function $f(x - \lfloor a\rfloor)$ is a constant value $f(0) - f(1)$.
Implied Differential Equations Conjecture
Any continuous solution to an equation constructed with implied derivatives of various orders is the same as the solution to the similarly constructed differential equation.
My preliminary thoughts:
[After the edit(s) on February 17, this answer is out of date.]
The Jump Series Theorem is definitely false, but since you gave it as a definition, we will consider the class of functions on which it is true. I won't say so explicitly in future lines, but from here on, I mean true/false with respect to this class.
The Jump Series Resolution Theorem is ill-defined, but the most reasonable interpretation that I can think of is true by definition.
The Integration Continuity Theorem is true by the theorem cited in the comments.
The Calculus Jump Theorem is, as best I can tell, meaningless because the floor function does not imply any sort of integration is happening; much less a constant of integration; even less a portion of such a constant, which is not a term I'm familiar with.
The Implied Integration and Differentiation Theorem is definitely false, but again I will from now on assume that we are only looking at the class of functions for which it is true.
The Jump Location Theorem is unambiguously true.
The Composite Floor Function Jump Sequence Theorem One is possible, but you don't explain what a jump sequence is. Based on the way you use it, it does not make sense for it to be a jump series, nor does it make sense to be the set of domain values where the function jumps. So I'm at a loss here.