Does $\exists$ a differentiable function $f:\mathbb{R}\to \mathbb{R}, f(x) \neq x+c$ s.t. every interval $(a,b)$ contains a point $p$ with gradient 1? Does there exist a differentiable function $f:\mathbb{R} \to \mathbb{R}, f(x) \neq x+c$ such that every interval $(a,b)$ contains a point $p$ with gradient $1$?
I would guess no, but I've no idea how to prove it.
Now that I think about it, isn't this question similar to: Does there exist a non-constant differentiable function $f:\mathbb{R} \to \mathbb{R}$, such that every interval $(a,b)$ contains a point $p$ with gradient $0$? Which I may or may not have seen on the site somewhere - can't remember. I guess the disproof would be something like: $f(x)$ has gradient $0$ almost everywhere $\implies f(x)$ has unbounded variation somewhere (e.g. in some interval) $\implies$ f(x) is not everywhere differentiable. I'm not that familiar with bounded variation other than having skim-read this thread once, but bounded variation may in fact not be necessary to answer this question, I've no idea.
Maybe the mean value theorem is more relevant here.
The answer is that yes, such a function exists.
The starting point is the Pompeiu Derivative..  To summarize Wikipedia, Pompeiu constructed an everywhere differentiable strictly increasing function $g:[0,1]\rightarrow \mathbb{R}$ whose derivative $g'(x)$ is $0$ on a dense subset of $[0,1]$.  Call this dense set $D$.
Now, let $h:\mathbb{R}\rightarrow (0,1)$ be your favorite diffeomorphism (e.g., you could pick $h(x) = \frac{1}{\pi}\arctan(x) + \frac{1}{2}$.)
Set $f(x) = g(h(x)) + x$.   I claim that $f$ fulfills your criterion.  By the chain rule, $f'(x) = g'(h(x))h'(x) + 1$.  For $x\in h^{-1}(D)$, $g'(h(x)) = 0$, so $f'(x) = 1$ for $x\in h^{-1}(D)$.  Note that since $h$ is a diffeomorphism, it is a homeomorphism, so $h^{-1}(D)\subseteq \mathbb{R}$ is dense.  Lastly, since $g$ is strictly increasing, $f(x) -x$ is strictly increasing, so is not constant.
Is c a constant in (a,b) interval? If yes, you can try to prove it by contradiction, like using the function f(x) = x^2 in interval ]-1, 1[. I guess that's what you want.(Sorry about formatting, I'm ne here)