I hope you don't mind, but I'm going to call $(x,y,z):= (a,b,c)$ and I'll denote the axial distance as $s=\sqrt{x^2+y^2}$ and I'll call $r=\sqrt{x^2+y^2+z^2}$ the radial distance, and finally, I'll call $l=\frac{L}{2}$, so that the wire goes from $-l$ to $l$ along the $z$-axis. With this notation, the $z$-component of the electric field is
\begin{align}
E_3&=\frac{\lambda}{4\pi \epsilon_0}\left[\frac{1}{\sqrt{s^2+(z-l)^2}}-
\frac{1}{\sqrt{s^2+(z+l)^2}}\right].
\end{align}
If you now wish look at a distant point in space, then you should be considering the limit $r\gg l$, rather than the limit $s\gg l$ as you have done (this is erroneous because now you're ignoring the $z$-effects of the field, which is why you're getting a zero field even though supposedly you're "far away" from an almost point-like source). More precisely, you should write things in terms of the quantity $\frac{l}{r}\ll 1$, and then calculate things in powers of this small quantity. So, we have
\begin{align}
E_3&=\frac{\lambda}{4\pi \epsilon_0}\left[\frac{1}{\sqrt{r^2-2zl+l^2}}-\frac{1}{\sqrt{r^2+2zl+l^2}}\right]\\
&=\frac{\lambda}{4\pi\epsilon_0}\cdot \frac{1}{r}\left[\frac{1}{\sqrt{1+\left(\frac{-2zl}{r^2}+\frac{l^2}{r^2}\right)}}-
\frac{1}{\sqrt{1+\left(\frac{2zl}{r^2}+\frac{l^2}{r^2}\right)}}\right]
\end{align}
Notice how in round brackets I have the term $\frac{\pm 2zl}{r^2}+\frac{l^2}{r^2}$, which in absolute value is bounded (using triangle inequality) by
\begin{align}
\left|\frac{\pm 2zl}{r^2}+\frac{l^2}{r^2}\right|&\leq 2\left|\frac{z}{r}\right|\cdot
\left|\frac{l}{r}\right| + \frac{l^2}{r^2}\\
&\leq 2\cdot 1\cdot \frac{l}{r}+\frac{l^2}{r^2}\\
&=\frac{2l}{r}+\left(\frac{l}{r}\right)^2
\end{align}
If $\frac{l}{r}\ll 1$ is a small quantity, then so is the above expression; more precisely what I'm saying is that the function $t\mapsto 2t+t^2$ has a limit of $0$ as $t\to 0$. Therefore, we now have a small quantity (namely $\frac{\pm 2zl}{r^2}+\frac{l^2}{r^2}$) in terms of which we can calculate the series expansion. I hope you recall that for small $|t|$, we have
\begin{align}
\frac{1}{\sqrt{1+t}}&=1-\frac{t}{2}+\frac{3}{8}t^2+\mathcal{O}(t^3)
\end{align}
Using this expansion above, we get
\begin{align}
E_3&=\frac{\lambda}{4\pi \epsilon_0}\cdot\frac{1}{r}\left[1-\frac{1}{2}\left(\frac{-2zl}{r^2}+\frac{l^2}{r^2}\right)+
\frac{3}{8}\left(\frac{-2zl}{r^2}+\frac{l^2}{r^2}\right)^2 +\mathcal{O}\left(\frac{-2zl}{r^2}+\frac{l^2}{r^2}\right)^3\right]\\
&-
\frac{\lambda}{4\pi \epsilon_0}\cdot\frac{1}{r}\left[1-\frac{1}{2}\left(\frac{2zl}{r^2}+\frac{l^2}{r^2}\right)+
\frac{3}{8}\left(\frac{2zl}{r^2}+\frac{l^2}{r^2}\right)^2 +\mathcal{O}\left(\frac{2zl}{r^2}+\frac{l^2}{r^2}\right)^3\right]\\
&=\frac{\lambda}{4\pi\epsilon_0}\cdot\frac{1}{r}\left[\frac{2zl}{r^2}+
\mathcal{O}\left(\frac{l}{r}\right)^3\right]\\
&=\frac{2\lambda l}{4\pi\epsilon_0}\cdot\frac{z}{r^3}+ \mathcal{O}\left(\frac{1}{r}\cdot \left(\frac{l}{r}\right)^3\right)\\
&=\frac{Q}{4\pi\epsilon_0}\frac{z}{r^3}+ \mathcal{O}\left(\frac{1}{r}\cdot \left(\frac{l}{r}\right)^3\right),\tag{$*$}
\end{align}
and so ignoring the higher order corrections, this first term is indeed the correct $z$-component of the electric field of due to a point charge $Q=2\lambda l$ located at the origin.
Remarks
Whenever you calculate some quantity in physics (electric field of some configuration/ gravitational potential of some collection of masses etc) and you wish to extract a certain limiting behavior, you should be careful to identify correctly the "small parameter". This has to be a dimensionless quantity, and you should then appropriately carry out the series expansion. In this case, you wanted the field far away from the source, and far-away here has to be interpreted as $r\gg l$, not just $s\gg l$. So it was in misidentifying the small parameter that led you to the erroneous $z$-behavior of the field.
Also, it is a good habit to get into, to keep track of the error terms so that you know just how good your approximation is. I know in physics texts, they don't usually pay attention to this sort of thing, but trust me, it'll make calculations clearer, and also make the range of validity of the approximation more apparent.
Extra Remarks (Can skip on a first read)
In fact, we can be slightly more precise with our error term. This is because there's a significant amount of cancellation going on. Let us write the power series (for $|t|<1$) as
\begin{align}
\frac{1}{\sqrt{1+t}}&=\sum_{n=0}^{\infty}c_nt^n,
\end{align}
for some constants $c_n$ (the first few terms being $c_0=1, c_1=-\frac{1}{2},
c_2=\frac{3}{8}$ etc). Then,
\begin{align}
E_3&=\frac{\lambda}{4\pi \epsilon_0 r}\sum_{n=0}^{\infty}c_n\left[\left(\frac{-2zl}{r^2}+\frac{l^2}{r^2}\right)^n-
\left(\frac{2zl}{r^2}+\frac{l^2}{r^2}\right)^n\right]\\
&=\frac{\lambda}{4\pi \epsilon_0 r}\sum_{n=0}^{\infty}c_n\left[\sum_{k=0}^n\binom{n}{k}\left(\frac{l^2}{r^2}\right)^{n-k}\cdot \left(\left(\frac{-2zl}{r^2}\right)^k-\left(\frac{2zl}{r^2}\right)^k\right)\right]\\
&=\frac{\lambda}{4\pi\epsilon_0 r}\sum_{n=0}^{\infty}c_n\sum_{\substack{0\leq k \leq n\\
\text{$k$ odd}}}\binom{n}{k}\left(\frac{l}{r}\right)^{2(n-k)}(-2)\left(\frac{2zl}{r^2}\right)^k,
\end{align}
where only the odd $k$'s contribute to the sum because when $k$ is even things cancel out. Now, when $n=0$, there are obviously no odd $k$'s between 0 and $n=0$. When $n=1$, we have $k=1$ is the only odd integer in that range, so we get a contribution
\begin{align}
n=1:\qquad \frac{\lambda}{4\pi \epsilon_0r}c_1\binom{1}{1}\cdot 1 \cdot (-2)\cdot \frac{2zl}{r^2}=\frac{\lambda}{4\pi\epsilon_0}\cdot \frac{2zl}{r^2}.
\end{align}
Now, starting from $n\geq 2$, we have several powers of $\frac{l}{r}$ (atleast 3 powers, since we're summing over odd $k$'s only, hence $k\geq 1$), and the improvement is that we can now clearly see that the error term also is of order $z$. So, in short,
\begin{align}
E_3&=\frac{\lambda}{4\pi\epsilon_0 r}\left[\frac{2zl}{r^2}+
\mathcal{O}\left(z\cdot \left(\frac{l}{r}\right)^3\right)\right]\\
&=\frac{Q}{4\pi\epsilon_0}\frac{z}{r^3}+\mathcal{O}\left(\frac{z}{r}\left(\frac{l}{r}\right)^3\right).\tag{$**$}
\end{align}
If you compare $(**)$ with $(*)$ above, you'll see that although the leading term obviously remains the same, we have been more precise with the error term in $(**)$, since we took into account the symmetry of the problem, and the resulting cancellations. This error term in $(**)$ shows us that when $z=0$, $E_3$ is indeed equal to $0$, as expected. So, $(**)$ gives us both the approximate field value when $r\gg l$, and also gives the correct behavior for when $\frac{z}{r}\ll 1$.
