While I was working the problem (and it was a pretty interesting problem), a complete answer appeared, but I think I'm still going to post all the results that I got so far. Not all of them are necessary for the final solution, but still interesting nonetheless.

Assume, there exists a differentiable function $f$ as described. Let $g\colon\mathbb{R}\rightarrow\mathbb{R}$ be the fixed point free function as described, that assigns a value $x\in\mathbb{R}$ the abscissa of the unique other point, where the tangent to $f$ at $(x,f(x))$ intersects the graph of $f$. Therefore $g(x)$ is the unique value, so that:
\begin{equation}
f(x)+f'(x)(g(x)-x)=f(g(x))
\Leftrightarrow
f'(x)=\frac{f(g(x))-f(x)}{g(x)-x}.
\end{equation}

**Conclusion**: If $g$ is continuous in $x$, then $f'$ is continuous in $x$.

**Proposition**: Changing $f(x)$ to $f(x)+\lambda x+\mu$ (with the same sought-after property) doesn't change $g(x)$.

**Corollary**: The conditions $g(g(x))=x$ and $f'(g(x))=f'(x)$ are equivalent.

*Proof*: $\Rightarrow$: Take the upper equation for $x$ and $g(x)$, combine them through $f(g(x))$ and use that $g$ doesn't have a fixed point to shorten $g(x)-x$. $\Leftarrow$: Directly use the upper equation and expand the fraction by $-1$. $\square$

**Lemma**: If $g$ is not continuous in $x$, then either the equivalent conditions of the upper corollary hold or $f'$ is not continuous in $x$.

Notice, that when $f$ is continuously differentiable, the first case always has to hold and that the backwards direction from the second case is the contraposition of the upper conclusion.

*Proof*: We prove the contraposition. Assume $f'(g(x))\neq f'(x)$ and $f'$ is continuous in $x$. Define $\widetilde{f}(y):=f(y)-f'(x)(y-x)-f(x)$ with $\widetilde{f}(x)=0$ and $\widetilde{f}'(x)=0$ as well as $\widetilde{f}(g(x))=0$ using the upper equation and proposition. (Particularly $\widetilde{f}$ doesn't have any other roots than $x$ and $g(x)$.) $\widetilde{f}'$ is continuous in $x$ and the assumption translates to $\widetilde{f}'(g(x))\neq 0$, so for every neighborhood $U$ of $g(x)$, $\widetilde{f}(U)$ is a neighborhood of $\widetilde{f}(g(x))$. Take a sequence $x_n\xrightarrow{n\rightarrow\infty}x$, then $\widetilde{f}(x_n)\xrightarrow{n\rightarrow\infty}\widetilde{f}(x)=0$ and $\widetilde{f}'(x_n)\xrightarrow{n\rightarrow\infty}\widetilde{f}'(x)=0$. The tangents $t_n(y)$ through $(x_n,\widetilde{f}(x_n))$ are given by $t_n(y)=\widetilde{f}(x_n)+\widetilde{f}'(x_n)(y-x_n)$. We have $t_n(g(x))\xrightarrow{n\rightarrow\infty}0$, so for all $n$ above a sufficient high boundary, $t_n(g(x))\in\widetilde{f}(U)$ and therefore $g(x_n)\in U$ (Notice, that a higher boundary may be necessary.), which implies $g(x_n)\xrightarrow{n\rightarrow\infty}g(x)$. $\square$

If $g$ is not continuous in $x$ and the first case of this lemma holds, then $g(x)-x$ changes signs between $x$ and $g(x)$, so there has to be another point where $g$ is not continuous between them, otherwise $\widetilde{g}$ would have a root or equivalently $g$ a fixed point due to the intermediate value theorem.

Assume $x<g(x)$ w.l.o.g. Because of the mean value theorem, there exists a $\xi\in(x,g(x))$ (which especially means $\widetilde{f}(\xi)\neq 0$) with $f'(\xi)=f'(x)\Leftrightarrow\widetilde{f}'(\xi)=0$ using the upper equation. Assume $\widetilde{f}$ is only positive in $(x,g(x))$ w.l.o.g. Let $\xi\in(x,g(x))$ be the value of the global maximum in $(x,g(x))$. It is unique since for two different maxima $\xi_1,\xi_2\in(x,g(x))$ with $\xi_1<\xi_2$ w.l.o.g., there would be a $\zeta\in(\xi_1,\xi_2)$ with $\widetilde{f}'(\zeta)=0$ and $\widetilde{f}(\zeta)<\widetilde{f}(\xi_1)=\widetilde{f}(\xi_2)$. Due to the intermediate value theorem, the horizontal tangent through $(\zeta,\widetilde{f}(\zeta))$ would intersect $\widetilde{f}$ also somewhere in $(x,\xi_1)$ and $(\xi_1,g(x))$. This also implies $g(\xi)\notin[x,g(x)]$. We are left with two cases for which we have to derive a contradiction:

**Case 1**: Assume $g(\xi)\in(-\infty,x)$, then $f>0$ in $(-\infty,0)$ and $f<f(\xi)$ in $(\xi,\infty)$.

**Case 2**: Assume $g(\xi)\in(g(x),\infty)$, then $f>0$ in $(g(x),\infty)$ and $f<f(\xi)$ in $(-\infty,x)$.

The argumentation for both cases is analogous and done in the other answer.