Say $P$ is a convex Euclidean polytope, where the origin is not contained in any bounding hyperplane containing a facet of $P$, with $n$ facets given by $\langle f_i , x\rangle = 1$ and $m$ vertices $v_j$ with $1\leq i\leq n$ and $1\leq j\leq m$. (That is, each facet equation has been multiplied through as needed to obtain $1$ on the right side.) Is it true that the polytope $P’$ with $m$ facets given by $\langle v_j , x\rangle = 1$ and $n$ vertices $f_i$ is combinatorially equivalent to the polar dual $P^*$ of $P$?
Yes. Your polytope $P'$ is exactly the polar $P^*$ of $P$ with respect to the standard unit sphere. Choosing another conic instead of the unit sphere will in general give another polar. See this wikpedia page for the two-dimensional case of polarity with respect to other conics. So your "Cartesian" dual corresponds to the unit sphere.
Polarity/duality of polytopes or more generally, convex bodies, is treated in many books on convexity. Unfortunately, many of them leave a lot of detail to the reader. I recommend Webster's Convexity, Brønsted's An Introduction to Convex Polytopes or Matousek's Lectures on Discrete Geometry.