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EDIT: While ljjpfx seems to be happy with this answer as his or her spaces are compact, Nate Eldredge asked in a comment whether there is a counterexample with both spaces Polish, which I think is an interesting question. I believe the following example should work. Consider the function$$f\colon\omega^\omega\to[0,1),\quad f(a_0,a_1,a_2,\dots)=0{.}\underbrace{1\cdots1}_{a_0}0\underbrace{1\cdots1}_{a_1}0\underbrace{1\cdots1}_{a_2}0\cdots,$$where the right-hand side is meant to be the binary representation. (I will write $0{.}1^{a_0}01^{a_1}0\cdots$ to simplify the notation.) It is easy to see that $f$ is a continuous bijection, while $f^{-1}$ is not continuous. Let $U$ be the open set$$\{a=(a_n)_{n\in\omega}\in\omega^\omega:\exists n\,a_n\text{ is even}\},$$I claim that $f(U)$ is not $G_\delta$. Assume for contradiction that $f(U)=\bigcap_nG_n$ for some open $G_n\subseteq[0,1)$. Note that $f(U)$ is the union of intervals of the form $[a2^{-k},(a+1)2^{-k})$, where $a,k\in\omega$, and $a$ has the binary representation $1^{a_0}01^{a_1}0\cdots01^{a_r}0$, where $a_i$ are odd for $i< r$, $a_r$ is even, and $k=\sum_{i\le r}(a_i+1)$. Since $G_n\supseteq f(U)$ is open, for every $n,k$ there exists $l_{n,k}$ such thatfor every $a=1^{a_0}01^{a_1}0\cdots01^{a_r}0$ as above. Note that if $a_r>0$, the end-point of this interval has the form $0{.}1^{a_0}01^{a_1}0\cdots01^{a_{r-1}}01^{a_r-1}01^{l_{n,k}+1}$. Choose $a=(a_n)\in\omega^\omega$ so that every $a_n$ is odd, and $a_{n+1}>l_{n,k_n}$, where $k_n=1+\sum_{i\le n}(a_i+1)$. Then $a\notin U$, but the description above implies that $f(a)\in\bigcap_nG_n$, a contradiction.
Yes, if $X$ is $\sigma$-compact: $X\smallsetminus E$ is $F_\sigma$, hence a countable union of compact sets. Since a continuous image of a compact set is compact (and therefore closed, in a Hausdorff space), $f(X\smallsetminus E)$ is also $F_\sigma$, hence $f(E)$ is $G_\delta$.