"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), 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 $1+U\ge\frac a{np+a}$. Also, if $U<-1/2$, then $1+U<1/2$, whence $\ln(1+U)<0$ and hence $|\ln(1+U)|\le \ln\frac{np+a}a$.
So, by
\begin{align*}
&E|\ln(1+U)\,1(U<-1/2)| \\
&\le\ln\frac{np+a}a E1(U<-1/2)=O(1/n^2), \tag{11}
\end{align*}
by (7) with $m=0$,
\begin{align*}
&E|\ln(1+U)\,1(U<-1/2)| \\
&\le\ln\frac{np+a}a E1(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.