For future students, here is a more general result:
Let $X$ and $Y$ be normed linear spaces, and let $B(X,Y)$ denote the collection of all bounded linear operators from $X$ to $Y$ endowed with the operator norm. Show that $B(X,Y)$ is a normed linear space, and $B(X,Y)$ is a Banach space whenever $Y$ is a Banach space. The vector operations in $B(X,Y)$ are defined pointwise, i.e. $(A+B)(x)=Ax+Bx$, and $(\alpha A)(x)=\alpha (Ax)$.
(Notice that in your case $X'=B(X,\mathbb{C})$ and $\mathbb{C}$ is a Banach space)
It is clear that linear operators form a linear space. To show that $B(X,Y)$ is a linear subspace, it is enough to show the closure to addition and scalar multiplication. But these follow easily from the properties of a norm (the fact that the operator norm satisfies all the properties of a norm for bounded functionals is an easy exercise that follows from properties of supremums in $[0, \infty)$) , namely for any $A,B \in B(X,Y)$ and $\lambda \in \mathbb{C}$
$$\|A+B\| \leq \|A\|+\|B\| < \infty$$
$$\|\lambda A\|=|\lambda| \cdot \|A\| < \infty$$
Thus, $B(X,Y)$ is a normed linear space.
Now assume that $Y$ is a Banach space. Let $\{A_i\}$ be a Cauchy sequence in $B(X,Y)$, i.e. $\forall \, \epsilon >0$, $\exists \, N \in \mathbb{N}$ such that $\forall \, m,n > N$, $\|A_n-A_m\|< \epsilon $. Let $x \in X$ be arbitrary. Let $\epsilon>0$ be arbitrary. If $x=0$, then
$$\|A_nx-A_mx\|=0<\epsilon.$$
If $x \neq 0$, choose $N$ such that $\|A_n-A_m\|< \frac{\epsilon}{\|x\|}$. Then by a property of the operator norm, $\forall \, m,n > N$,
\begin{equation}
\begin{split}
\|A_nx-A_mx\|
& = \|(A_n-A_m)x\|\\
& \leq \|(A_n-A_m)\| \cdot \|x\|\\
& < \frac{\epsilon}{\|x\|} \cdot \|x\|\\
& = \epsilon\\
\end{split}
\end{equation}
Thus, in both cases $\{A_nx\}$ is a Cauchy sequence in $Y$. Since $Y$ is a Banach space, it is convergent to some element in $Y$. Call that element $Ax$, i.e.
$$\lim_{n \rightarrow \infty} A_nx=Ax$$
Since $x$ was arbitrary, $Ax$ is defined for any $x \in X$. Thus, $A$ is a map from $X$ to $Y$ defined by $x \rightarrow Ax$. We need to show that $A$ is linear, bounded, and $A_n \xrightarrow{n \rightarrow \infty} A$ in the operator norm. Notice that $A$ is linear, since by linearity of $A_n$ we get that for any $x_1, x_2 \in X$, $\lambda \in \mathbb{C}$,
\begin{equation}
\begin{split}
A(x_1+x_2)
& = \lim_{n \rightarrow \infty} A_n(x_1+x_2)\\
& = \lim_{n \rightarrow \infty} (A_nx_1+A_nx_2)\\
& = \lim_{n \rightarrow \infty} A_nx_1+\lim_{n \rightarrow \infty} A_nx_2\\
& = Ax_1+Ax_2\\
\end{split}
\end{equation}
\begin{equation}
\begin{split}
A(\lambda x_1)
& = \lim_{n \rightarrow \infty} A_n(\lambda x_1)\\
& = \lim_{n \rightarrow \infty} \lambda \cdot A_nx_1\\
& = \lambda \lim_{n \rightarrow \infty} A_nx_1\\
& = \lambda\cdot Ax_1\\
\end{split}
\end{equation}
Now recall that Cauchy sequences are bounded. Thus, $\forall \, n$, $\|A_n\|<C$ for some $C \in \mathbb{R}$. Using this fact, we can see that $A$ is bounded, since by continuity of a norm:
\begin{equation}
\begin{split}
\|A\|
& =\sup_{\|x\| \leq 1} \|Ax\|\\
& =\sup_{\|x\| \leq 1} \|\lim_{n \rightarrow \infty} A_nx\|\\
& =\sup_{\|x\| \leq 1} \lim_{n \rightarrow \infty} \|A_nx\|\\
& =\sup_{\|x\| \leq 1} \limsup_{n \rightarrow \infty} \|A_nx\|\\
& \leq \sup_{\|x\| \leq 1} \limsup_{n \rightarrow \infty} \Big(\|A_n\|\cdot \|x\|\Big)\\
& \leq \sup_{\|x\| \leq 1} C \cdot \|x\|\\
& = C \sup_{\|x\| \leq 1} \|x\|\\
& \leq C \\
\end{split}
\end{equation}
Finally, we want to show that $A_n \xrightarrow{n \rightarrow \infty} A$ in the operator norm. Let $\epsilon > 0$ be arbitrary. Recall that for an arbitrary $x \in X$, we have
$$\|A_nx-A_mx\| \leq \|(A_n-A_m)\| \cdot \|x\|$$
Since $\{A_n\}$ is Cauchy, choose $N$ big enough such that for all $n,m \geq N$, $\|(A_n-A_m)\| < \epsilon$. Then the above inequality turns into
$$\|A_nx-A_mx\| \leq \epsilon \cdot \|x\|$$
Now by continuity of a norm, we can take limit on both sides as $m$ goes to infinity to obtain
$$\|A_nx-Ax\| \leq \epsilon \cdot \|x\|$$
Now taking supremum on both sides over all $x$ such that $\|x\| \leq 1$ yields
$$\sup_{\|x\| \leq 1}\|A_nx-Ax\| \leq \epsilon$$
But this is equivalent to saying that for all $n \geq N$,
$$\|A_n-A\| \leq \epsilon$$
And since $\epsilon$ was arbitrary, this implies that
$$A_n \xrightarrow{n \rightarrow \infty} A$$
in the operator norm. Thus, we conclude that $B(X,Y)$ is a Banach space.
The answer is "No". Take your favorite infinite-dimensional Banach space $Y$ and choose (alas, this requires AC, so if you are not a believer, stop reading here) some discontinuous linear functional $\psi$ on $Y$. Let $X=Ker(\psi)$ with the norm inherited from $Y$. Assume now that $A:X\to X$ has norm less than $1$. Then it extends by continuity to $Y$ and enjoys the property $AX\subset X$, so $\psi\circ A$ vanishes on $X=Ker(\psi)$. We want to show that if $y-Ay=x\in X$, then necessarily $y\in X$. Assume not. Then $\psi(y)\ne 0$ and $(\psi\circ A)(y)=\psi(y)$. Hence the linear functionals $\psi\circ A$ and $\psi$ coincide on the entire space $Y$, so $\psi(z-Az)=0$ for all $z\in Y$, which is absurd because $z-Az$ is just an arbitrary element of $Y$ and $\psi$ is certainly not identically $0$.
The trick is, of course, that it is not so easy for a continuous linear operator to preserve the kernel of a discontinuous linear functional, so the operators acting from $X$ to $X$ are rather few in a certain sense, which makes the Neumann series convergence condition vacuous exactly when it starts getting interesting.
Best Answer
To show that the space is complete we need to show that every Cauchy sequence in the space converges to a limit in the space. Here is a counter-example.
Consider the sequence $\lbrace S_n\rbrace$ defined by
$S_n=(1,\frac{1}{2},\frac{1}{3},..,\frac{1}{n},0,0,...)$. The limit of this sequence is $S=(1,\frac{1}{2},\frac{1}{3},..,\frac{1}{n},\frac{1}{n+1},...)$.
Now the sum $\sum^\infty_{n=1}\frac{1}{n^2}=\frac{\pi^2}{6}$ is convergent, so the partial sums $\sum^n_{i=1}\frac{1}{i^2}$ form a Cauchy sequence (in $\mathbb{R}$).
It follows that the sequence $\lbrace S_n\rbrace$ is Cauchy in $||.||_2$, since if $m>n$, $$||S_m-S_n||_2=\sqrt{(\frac{1}{n+1})^2+(\frac{1}{n+2})^2+..+(\frac{1}{m})^2}<\sqrt{\frac{\pi^2}{6}-\sum^\infty_{n=1}\frac{1}{n^2}}\to0$$ as $n\to\infty$.
For each $n$, $||S_n||_1=\sum^n_{i=1}\frac{1}{i}$ is a finite sum, and so $S_n \in$ l$^1$. However the sum $\sum^\infty_{n=1}\frac{1}{n}$ is not convergent. Hence $||S||_1$ is infinite and so $S\notin$ l$^1$.
The sequence $\lbrace S_n\rbrace$ is therefore a Cauchy sequence in $($l$^1,||.||_2)$ with no limit in $($l$^1,||.||_2)$. The space is therefore not complete.