A jump continuous function is defined as a function such that all the lateral limits of it points are well-defined. I think that the statement of the title is true, can you check my proof please?
We previously knows that a jump continuous function have, at most, a countable number of discontinuities, and this subset of discontinuities is a set of isolated points.
Then what define jump continuous functions is it continuity at any lateral neighborhood (if exists) of any (limit) point of it domain. Let define a left-neighborhood of some point $x$ as the interval $(x-\delta,x)$ for some $\delta>0$, and a right-neighborhood of $x$ as the interval $(x,x+\delta)$.
Suppose that $f$ have a discontinuity at some $g(x)$. I claim that exists lateral neighborhoods at the sides of $x$ such that $f\circ g$ is continuous. Proof: there are two cases:
If $g$ is constant in some lateral neighborhood then the image of this neighborhood is a point, hence $f\circ g$ is continuous in this neighborhood.
If $g$ is not constant at some lateral neighborhood then the image of this neighborhood is an interval. Because $f$ is jump continuous then for any point in that interval the function is continuous at it laterals, hence there is a lateral neighborhood of any point where the function is continuous, what imply that $f\circ g$ is continuous in this neighborhood.
By the other side if $g$ have a discontinuity at $x$ then $g$ is continuous at it lateral neighborhoods, hence, by the same reasons shown above, exists lateral neighborhoods of $x$ such that $f\circ g$ is continuous.
Hence for any point of $\mathrm{dom}(f)$ exists lateral neighborhoods where $f\circ g$ is continuous, hence $f\circ g$ is jump continuous.$\Box$.