"And I'm only interested in the case when the Taylor series is convergent for all values of $x$." -- There is no such Taylor series for $\ln$ (which is only defined on $(0,\infty)$). 

However, to get the desired $O(1/n^2)$, you do not need a convergent series; you do not need any series at all. Instead, you need the Taylor expansion 
\begin{equation}
	\ln(1+u)=u-u^2/2+u^3/3+O(u^4) \tag{1}
\end{equation}
for $u\ge-1/2$ (say), with a universal constant in $O(u^4)$. 

Indeed, letting 
\begin{equation}
	U:=\frac{X-np}{np+a}, \tag{2}
\end{equation}
for all real $a>0$ you can write 
\begin{align*}
	&E\ln(X+a)-\ln(np+a) \\
	&=E\ln(1+U) \\ 
	&=E\ln(1+U)1(U\ge-1/2) \\ 
	&+E\ln(1+U)1(U<-1/2).  \tag{3}
\end{align*}
By (1),  
\begin{align*}
	&E\ln(1+U)\,1(U\ge-1/2) \\ 
	&=E(U-U^2/2+U^3/3+O(U^4))\,1(U\ge-1/2) \\ 
	&=E(U-U^2/2+U^3/3)+O(EU^4) \\ 
	&-E(U-U^2/2+U^3/3)1(U<-1/2). \tag{4}
\end{align*}
Now we are going to use (say) Rosenthal's inequality (see e.g. formula (7) in [this paper][1]), which implies 
\begin{equation}
	EU^4=O(1/n^2) \tag{5}
\end{equation}
and 
\begin{equation}
	EU^6=O(1/n^3); \tag{6}
\end{equation}
here in what follows, the constants in $O(\cdot)$ may depend only on $p$ (and, in (11) and (12), also on $a$). 
By (6), for $m=0,1,2,3$ we have $|U|^m\,1(U<-1/2)|\le U^6(1/2)^{m-6}$ and hence 
\begin{equation}
	|EU^m\,1(U<-1/2)|\le EU^6(1/2)^{m-6}=O(1/n^3). \tag{7}
\end{equation} 
Also, 
\begin{equation}
	EU=0,\quad EU^2=\frac{npq}{(np+a)^2} \tag{8}
\end{equation}
(where $q:=1-p$), 
$E(X-np)^3=n(pq^3-qp^3)=O(n)$ and hence 
\begin{equation}
	EU^3=O(1/n^2). \tag{9}
\end{equation}
By (4), (5), (7), (8), (9), 
\begin{align*}
	&E\ln(1+U)\,1(U\ge-1/2) \\ 
	&=E(U-U^2/2+U^3/3)+O(1/n^2) \\ 
	&=-\frac{npq}{2(np+a)^2}+O(1/n^2). \tag{10}
\end{align*}
By (2) and the obvious inequality $X\ge0$, we have $1+U\ge\frac a{np+a}$. So, by (7) and (6),
\begin{align*}
	&E|\ln(1+U)\,1(U<-1/2)| \\ 
	&\le\ln\frac{np+a}a\,EU^6(1/2)^{-6}=O(1/n^2). \tag{11}
\end{align*}
Now (3) and (10) yield 
\begin{align*}
	&E\ln(X+a)=\ln(np+a)-\frac{npq}{2(np+a)^2}+O(1/n^2), \tag{12}
\end{align*}
as desired. 

[1]: https://projecteuclid.org/journals/annals-of-probability/volume-43/issue-5/Exact-Rosenthal-type-bounds/10.1214/14-AOP942.full