I have the following setup: denote $\mathcal{B}$ the Borel $\sigma$-algebra on $\mathbb{R}$ and $\lambda$ the Lebesgue measure on $\mathbb{R}$. Let $\mathcal{M}$ be the $\sigma$-algebra generated by $\{A_1 \times A_2: A_1, A_2 \in \mathcal{B}\}$ on $\mathbb{R}^2$.
If $S \in \mathcal{M}$, then we want to show that $\{x: (x, c+x) \in S\}$ is in $\mathcal{B}$ for all $c \in \mathbb{R}$.
I've observed that $\mathcal{M}$ is a sub-$\sigma$-algebra of the product $\sigma$-algebra for $\lambda \times \lambda$. I only see that $\{x: (x, c+x) \in S\}$ represents the $x$-values for which the lines $y=x+c$ intersects $S$. However, I don't know to where to go from here to show that it is Borel.
The basic idea with most of these exercises is two notice that the sets in question are defined by some equations from which you can build a function.
In that sense consider the function $f:\mathbb R\to\mathbb R^2,\ x\mapsto (x,x+c)$. This function is clearly continuous as it is component-wise continuous.
Now, as with open sets, Borel sets are sent back to Borel sets by continuous functions. This is, if $f:X\to Y$ is continuous, then if $B\subseteq Y$ is Borel, then $f^{-1}[B]\subseteq X$ is Borel.
Your set is precisely the inverse image of $S$:
$$f^{-1}[S]=\{x:\ f(x)\in S\}=\{x:\ (x,x+c)\in S\}.$$
So as $S$ is a Borel set, then $f^{-1}[S]$ is also a Borel set.