Let $S\subseteq [0,N]$ for some $N\geq 1$ be a measurable set with positive measure.
The first question is to show that there is $x\in S$ and $t>0$ so that $x+t\in S$ and $x+2t\in S$. To do this, my idea is reformulating the problem: $$ I:=\int_0^N \int_0^{N} 1_{S}(x)1_{S}(x+t)1_S(x+2t)dt\,dx $$ Then if $I>0$, we could conclude our claim. I intend to use either the Lebesgue differentiation theorem or properties of convolutions. How could I proceed?
More generally, I want to consider $P(t)$ to replace the "$2t$" above, where $P(0)=0$ and $P$ is continuous at $0$ (If necessary, you may assume that $P$ is real analytic near $0$), so that the problem becomes showing $$ \int_0^N \int_0^{N} 1_{S}(x)1_{S}(x+t)1_S(x+P(t))dt\,dx>0. $$
Any suggestions?
This answer is inspired by this.
Lemma 1 : Let $T \subset [-1,1]$ a measurable set with measure $\lambda(T)>1$. Then $T \cap (-T)$ is non-empty.
Indeed, let us denote by $T^\complement$ the set $[-1,1] \setminus T$. Then $$ \begin{align} \lambda(T\cap(-T))&=2-\lambda(T^\complement \cup(-T)^\complement) \\ &\geq2-\lambda(T^\complement)-\lambda((-T)^\complement) \\ &=2-(2-\lambda(T))-(2-\lambda(T)) \\ &=2(\lambda(T)-1) \\ &>0. \end{align} $$
Proof of your first claim : Now lets us get back to our set $S \subset \mathbb{R}$ of positive measure. By Lebesgue's density theorem, for almost every $x \in S$, we have $$\lim_{r\to 0} \frac{\lambda([x-r,x+r]\cap S)}{2r} =1. $$
Since $S$ has positive measure, we can take $x_0 \in S$ that satisfies the above condition, and $r>0$ such that $$\lambda([x_0-r,x_0+r]\cap S)>r. $$
Using the lemma, one can see that $([x_0-r,x_0+r]\cap S)\cap([x_0-r,x_0+r]\cap (2x_0-S))$ is non-empty, and a fortiori $S \cap (2x_0-S)$ is, which means that there exists $t\in S \cap (2x_0-S)$ and then $(t,x_0,2x_0-t)$ is an arithmetic progression contained in $S$.
Disclaimer : I'm not too sure about the next part, some exterior advice is welcome.
Lemma 2 : Let $P \colon [-1,1]\to [-1,1]$ be measurable with $P(0)=0$ and $P$ continuous at $0$. I will also assume that there is no neighborhood of $0$ such that $P$ is almost everywhere constant in it. There exists $\varepsilon>0$ such that, if $T \subset [-1,1]$ is a measurable set with measure $\lambda(T)>2-\varepsilon$, then $P^{-1}(T)\cap(-T)$ is non-empty.
We compute once again : $$ \begin{align} \lambda(P^{-1}(T)\cap(-T))&=2-\lambda(P^{-1}(T)^\complement \cup (-T)^\complement) \\ &\geq 2-(2-\lambda(P^{-1}(T)))-(2-\lambda(T)) \\ &=\lambda(P^{-1}(T))-(2-\lambda(T)). \end{align} $$
Thanks to the hypotheses on $P$, there exists $\varepsilon>0$ such that $\lambda(P^{-1}(T))>0$ as soon as $\lambda(T)>2-\varepsilon$. If needed, take a lower value $\varepsilon'<\varepsilon$ so the term $2-\lambda(T)$ can be as low as we want, and we get the lemma.
Proof of the second claim : Let $S\subset \mathbb{R}$ be of positive measure. Let $P$ satisfying the above conditions. As before, we can find $x_0$ and $r$ such that $$\lambda([x_0-r,x_0+r]\cap S)>r(2-\varepsilon). $$ By the lemma we have a point $t_0\in P^{-1}(-x_0+S)\cap(x_0-S)$, and letting $t=x_0-t_0$ we have a progression $(t,x_0,P(x_0-t)-x_0)$ with the three terms belonging in $S$.
Edit : Restating the last part so that it matches the question. Let $P,Q$ be two functions that satisfy the right conditions. Then we have a progression $$(u,x_0,Q(x_0-u)-x_0) $$ where the terms belong in $S$. Now let $x=u$, $t=x_0-u$, and $P(X)=Q(X)-X$. The progression rewrites as $$(x,x+t,x+P(t)). $$