Adjoint operator, bijective

This is not true in general. Let $1<p<q<\infty$. Consider bijective bounded linear operator $$ I:(\ell_p,\Vert\cdot\Vert_p)\to(\ell_p,\Vert\cdot\Vert_q):x\mapsto x $$ which is set theoretic identity. Its adjoint is $$ I^*:(\ell_p,\Vert\cdot\Vert_q)^*\to(\ell_p,\Vert\cdot\Vert_p)^*:f\mapsto f\circ I $$ Assume it is bijective. Since dual spaces are always complete, then $I^*$ is a bijection between Banach spaces. By open mapping theorem $I^*$ is an isomorphism. Note that $$ (\ell_p,\Vert\cdot\Vert_p)^*\cong(\ell_q,\Vert\cdot\Vert_q) $$ $$ (\ell_p,\Vert\cdot\Vert_q)^* \cong\left(\operatorname{completion}(\ell_p,\Vert\cdot\Vert_q)\right)^* \cong(\ell_q,\Vert\cdot\Vert_q)^* \cong(\ell_p,\Vert\cdot\Vert_p) $$ so using that $I^*$ is an isomorphism we see that $(\ell_p,\Vert\cdot\Vert_p)$ and $(\ell_q,\Vert\cdot\Vert_q)$ are isomorphic via some operator $J$. But this is impossible, so we get a contradiction. Thus $I^*$ is not bijective, though $I$ is bijective.

On the other hand, if you assume that $X$ and $Y$ are Banach spaces then you have an equivalence $$ A\text{ is bijective}\Longleftrightarrow A'\text{ is bijective} $$ For proof see this and this answers.