Let $f \in \mathcal{S}(\mathbb{R}^2)$. Then do there exists inequalities of the form $$\lVert f(x, A(x,y))\rVert_{L^p_x(\mathbb{R})} \lesssim_{A, p,q} \lVert f(x,y) \rVert_{L^q(\mathbb{R}^2)} $$ where $A : \mathbb{R}^2 \to \mathbb{R} $ is linear?
An often studied case is when $ f$ is separable: That is if $ f(x, y) = K(x) g(-y)$ then we have $$ f(x , x - y) = K(x) g( y - x)\quad \text{ implying} $$ $$ \lVert f (x, x - y)\rVert_{L^1_x} = \lVert K(x) g(y - x ) \rVert_{L^1_x} =| K \star g(y)| \leq \lVert K \star g \rVert_{L^\infty} \leq \lVert K \rVert_{L^2} \lVert g \rVert_{L^2} = \lVert f \rVert_{L^2}. $$ So are there similar results for some more general assumptions on $f?$
Let $A(x, y)=ax+by$. Up to a rotation and a dilation of $\mathbb R^2$ we can assume $a=1, b=0$. Thus, you are asking whether $$\tag{1} \left(\int_{-\infty}^\infty |f(x, 0)|^p\, dx\right)^\frac1p\lesssim \left(\int_{\mathbb R^2} |f(x, y)|^q\, dxdy\right)^\frac1q.$$ As you mention in comments to this answer, this is not possible. Take $f\in L^q(\mathbb R^2)$. You can always rescale it as $f_\lambda(x, y)=f(x, \lambda y)$. The left-hand side of (1) is left unchanged, but the right-hand side is not, and you can make it arbitrarily small. Thus, (1) cannot hold with a constant that is independent on $f$.
In technical jargon you are asking whether an $L^q(\mathbb R^2)$ function can have an $L^p(\mathbb R)$ trace. This is not possible, as we just saw. You need to require more regularity on $f$, which will rule out the scaling counterexample. For more information on this subject, check https://en.wikipedia.org/wiki/Trace_operator, or look up "Trace theorems" in your favorite book in PDE.
PREVIOUS VERSION OF THIS ANSWER. Here I erroneously refer to a conjectured inequality $$\tag{2} \lVert f(x, A(x, y))\rVert_{L^p_{x, y}(\mathbb R^2)}\lesssim \lVert f \rVert_{L^q(\mathbb R^2)}.$$
Provided $b\ne 0$, the left-hand side of (2) is actually equivalent to $\lVert f \rVert_p$. Indeed, letting $(x', y')=(x, A(x, y))$ we see that $dx'dy'=|b|dxdy$, so $$ \int_{\mathbb R^2} |f(x, A(x, y))|^p\, dxdy=\frac{1}{|b|} \int_{\mathbb R^2} |f(x', y')|\, dx'dy'.$$
Therefore, you are essentially asking whether $\lVert f\rVert_p\lesssim \lVert f\rVert_q$, which is only possible for arbitrary $f$ in the trivial case $p=q$.