Under some set-theoretic assumptions the answer to this question is negative.

Namely, if there exists a $Q$-set $X$, then $X$ is a $\sigma$-set which is Borel isomorphic to a (hereditarily normal compact) topological space $Y$ which is not a $\sigma$-space.

A topological space $X$ is called a $Q$-*space* if each subset of $X$ is of type $F_\sigma$. A $Q$-space is called a $Q$-*set* if $X$ is a subset of the real line. It is well-known that uncountable $Q$-sets exist under MA+$\neg$CH (more precisely, under MA each subset $A\subset\mathbb R$ of cardinality $|A|<\mathfrak c$ is a $Q$-set).

Now take any uncountable $Q$-set $X$ and let $Y$ be the one-point compactification of a discrete space of cardinality $|X|$. Each subset of $Y$ is Borel (more precisely, open or closed). Then any bijective map $f:X\to Y$ is a Borel isomorphism. But $Y$ is not a $\sigma$-space since for the unique non-isolated point $y$ of $Y$ the singleton $\{y\}$ is not of type $G_\delta$ in $Y$.

But this answer is a bit unfair. It would be much interesting to know if a $\sigma$-set can be Borel isomorphic to a subset of the real line which is not a $\sigma$-set.

This question is related to an old open problem of Miller who asked in 1979 if for any ordinals $2\le \alpha<\beta<\omega_1$, a $Q_\alpha$-set $A$ and a $Q_\beta$-set $B$ we have $|A|<|B|$. A counterexample to this Miller's problem for $\alpha=2$ would give a $Q$-set (and hence $\sigma$-set) $A$ which is Borel isomorphic (by any bijection) to a $Q_\beta$-set which is not a $\sigma$-set. But this problem remains open since 1979, so seems to be difficult.

Concerning the problem of preservation of $\sigma$-sets by Borel isomorphisms, let us remark that Serpinski sets (which are $\sigma$-sets) are preserved by Borel isomorphisms.