Write $[a,b]:=\{a,a+1,a+2,\ldots,b-1,b\}$ for all $a,b\in\mathbb{Z}$ such that $a\leq b$ (if $a>b$, then $[a,b]:=\emptyset$). For the set $A$ to exist, $k\geq 4$ must be true.
First, we deal with the case $k=4$, whence $|A+A|=9$. We may suppose without loss of generality that $A=\{0,a,b,c\}$ with $0<a<b<c$. Then,
$$A+A=\{0,a,2a,a+b,2b,b+c,2c\}\cup\{b,c\}\cup\{a+c\}\,.$$
Since $S:=\{0,a,2a,a+b,2b,b+c,2c\}$ already has $7$ distinct elements, the two extra elements can only come from two of the three expressions $b$, $c$, and $a+c$.
Case I: $A=S\cup\{b,c\}$. Then, $a+c$ must be equal to $2b$. Therefore, $a$, $b$, and $c$ form an arithmetic progression. That is, $A=\{0,a,a+d,a+2d\}$ for some positive integers $a$ and $d$. This gives
$$A+A=\{0,a,a+d,a+2d,2a,2a+d,2a+2d,2a+4d\}\,,$$
but as $|A+A|=9$, we need $d\notin\left\{a,\dfrac{a}{2}\right\}$.
Case II: $A=S\cup\{a+c,b\}$. Then, $c$ must belong in $S$. Hence, $c\in\{2a,a+b,2b\}$.
If $c=2a$, then $A=\{0,a,b,2a\}$, so
$$A+A=\{0,a,b,2a,a+b,2b,3a,2a+b,4a\}\,.$$
Note that we need $b<2a$ and $b\neq \dfrac{3a}{2}$.
If $c=a+b$, then $A=\{0,a,b,a+b\}$, so
$$A+A=\{0,a,b,2a,a+b,2a+b,2b,a+2b,2a+2b\}\,.$$
We require $b\neq 2a$.
If $c=2b$, then $A=\{0,a,b,2b\}$, so
$$A+A=\{0,a,b,2a,a+b,2b,a+2b,3b,4b\}\,.$$
We require $b\neq 2a$.
Case III: $A=S\cup\{c,a+c\}$. Then, $b$ must belong in $S$, making $b=2a$ the only possibility. Thus, $A=\{0,a,2a,c\}$ with $c>2a$, whence
$$A+A=\{0,a,2a,3a,4a,a+c,2a+c,2c\}\,.$$
Now assume that $k\geq 5$. As you claimed (I did not read the book, so I hope there is no mistake at this step), we may assume that
$$A=[0,c-1]\cup[c+1,d-1]\cup[d+1,k+1]$$
for some $c,d\in[0,k+1]$ such that $c<d$. Observe that $(c,d)$ cannot be equal to $(0,1)$, $(k,k+1)$, or $(0,k+1)$ (otherwise, $A$ is an arithmetic progression of length $k$, so that $A+A$ has $2k-1$ elments).
Case I: $c=0$. Then, $2\leq d\leq k$ and $A=[1,d-1]\cup[d+1,k+1]$. This gives
$$A+A=[2,2d-2]\cup[d+2,d+k]\cup[2d+2,2k+2]\subseteq [2,2k+2]\,.$$
Because $[2,2k+2]$ has exactly $2k+1$ elements, we get $A+A=[2,2k+2]$. This shows that $2d-2\geq (d+2)-1$ and $d+k\geq (2d+2)-1$. That is, $3\leq d\leq k-1$.
Case II: $d=k+1$. Using a similar argument as Case I, we obtain $2\leq c\leq k-2$.
Case III: $d=c+1$ with $1\leq c\leq k-1$. Then, $A=[0,c-1]\cup[c+2,k+1]$. That is,
$$A+A=[0,2c-2]\cup[c+2,c+k]\cup [2c+4,2k+2]\,.$$
Hence, $A+A\subseteq [0,2k+2]$ and $[0,2k+2]\setminus(A+A)$ contains two elements.
If $c=1$, then $A+A=\{0\}\cup[3,k+1]\cup[6,2k+2]=\{0\}\cup [3,2k+2]$ (since $k\geq 4$). Therefore, $A+A$ has $2k+1$ elements.
If $c=k-1$, then $A+A=[0,2k-1]\cup\{2k+2\}$ (since $k\geq 4$). Therefore, $A+A$ has $2k+1$ elements.
If $2\leq c\leq k-2$, then the two elements of $[0,2k+2]\setminus(A+A)$ must be $2c-1$ and $c+k+1$, which have to be less than $c+2$ and $2c+4$, respectively. This means $2c-1\leq (c+2)-1$ or $c\leq 2$, and $c+k+1\leq (2c+4)-1$ or $c\geq k-2$ That is, $k-2\leq c\leq 2$. Hence, $k=4$, which is a contradiction.
Case IV: $1\leq c\leq k-2$ and $c+2\leq d\leq k$. We then see that
$$\begin{align}A+A&=\big[0,2c-2\big]\cup[c+1,c+d-2]\cup [d+1,c+k]\\&\phantom{aaaaa}\cup [2c+2,2d-2]\cup[c+d+2,d+k]\cup[2d+2,2k+2]\,,\end{align}$$
which is a subset of $[0,2k+2]$. Since $|A+A|=2k+1$, the set $B:=[0,2k+2]\setminus(A+A)$ has two elements.
If $c=1$, then $$A+A=\{0\}\cup[2,d-1]\cup[d+1,k+1]\cup[4,2d-2]\cup[d+3,d+k]\cup[2d+2,2k+2]\,.$$ Thus, $1\in B$. If $4\leq d\leq k-1$, then $$[2,d-1]\cup[d+1,k+1]\cup[4,2d-2]\cup[d+3,d+k]\cup[2d+2,2k+2]=[2,2k+1]\,,$$ which leads to a contradiction. Hence, $d=3$ or $d=k$. If $d=3$, then
$$A+A=\{0,2\}\cup[4,2k+2]\,,$$
which has $2k+1$ elements. If $d=k$, then
$$A+A=\{0\}\cup[2,2k]\cup\{2k+2\}\,,$$
which also has $2k+1$ elements.
If $d=k$, then similarly to the former subcase, we can see that either $c=1$ or $c=k-2$. The case $(c,d)=(1,k)$ has already been covered in the previous subcase. In the case $(c,d)=(k-2,k)$, we have $A=[0,k-3]\cup\{k-1,k+1\}$, so
$$A+A=[0,2k-2]\cup\{2k,2k+2\}\,,$$
which has $2k+1$ elements.
If $2\leq c\leq k-3$ and $c+2\leq d\leq k-1$, then
$$[0,2c-2]\cup[c+1,c+d-2]\cup[d+1,c+k]=[0,c+k]$$
and
$$[c+d+2,d+k]\cup[2d+2,2k+2]=[c+d+2,2k+2]\,.$$
However, $[0,c+k]\cup[c+d+2,2k+2]=[0,2k+2]$. This subcase is impossible.
Here is the summary of all possible sets $A\subseteq \mathbb{Z}$, where $k\geq 4$ is an integer, such that $|A|=k$ with $|A+A|=2k+1$. The set $A$ must be an affine transformation of one of the following sets $A'$ (that is, $A=\{px+q\,|\,x\in A'\}$ for some fixed $p\in\mathbb{Z}_{\neq 0}$ and $q\in\mathbb{Z}$):
$k=4$ and $A':=\{0,a,a+d,a+2d\}$, where $d\in\mathbb{Z}_{>0}\setminus\left\{a,\dfrac{a}{2}\right\}$ and $\gcd(a,d)=1$;
$k=4$ and $A':=\{0,a,b,2a\}$, where $b\in\mathbb{Z}_{>0}\setminus\left\{\dfrac{a}{2},\dfrac{3a}{2},2a\right\}$ and $\gcd(a,b)=1$;
$k=4$ and $A':=\{0,a,b,a+b\}$, where $b\in\mathbb{Z}_{>a}\setminus\{2a\}$ and $\gcd(a,b)=1$;
$k\geq 5$ and $A':=[0,c-1]\cup[c+1,k]$, where $2\leq c\leq k-2$;
$k\geq 5$ and $A':=\{0\}\cup[2,k+1]$;
$k\geq 5$ and $A':=\{0,2\}\cup[4,2k+2]$;
$k\geq 5$ and $A':=\{0\}\cup[2,k-1]\cup\{k+1\}$.
