Let $A\in\mathcal{L}(X,Y)$, where $X,Y$ are normed vector spaces. Define the adjoint operator $$\begin{array}{ll} A^{\prime}\ : & Y^{\prime}\rightarrow X^{\prime},\\ & G \mapsto A^{\prime}(G)\ =\ G\circ A. \end{array}$$ Its easy to show that $A^{\prime}\in\mathcal{L}(Y^{\prime}, X^{\prime})$ and $\|A\| = \|A^{\prime}\|$.

Now, suppose that $A$ is bijective, then show that $A^{\prime}$ is also bijective.

First at all, I have proven that $[\mbox{Im}(A)]^{\circ} = \mbox{Ker}(A^{\prime})$, and then $\mbox{Ker}(A^{\prime}) = \{O\}$. Therefore $A^{\prime}$ is injective. My problem is that I don't know how I can show that $A^{\prime}$ is onto.

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.
I don't think the statement is true when the spaces are not complete. Let $X$ be the space of differentiable functions on $[-1,1]$ with the $C^1$ norm, and let $Y$ be the same set of functions, but equipped with the $C^0$ norm. Let $A$ be the identity map. It is certainly continuous, since $\|f\|_0 \leq \|f\|_1$.
Now, let $T(f) = f'(0)$. This is in $X'$, but is not the image of any operator under $A'$. Suppose $S: Y \rightarrow \mathbb{R}$ were such an operator. Then $S(f) = T(f)$. But this means that differentiation at $0$ is a continuous map on $Y$. But there are functions bounded by 1 that have arbitrarily large derivatives at 0.