Excellent question indeed. The quick answer is that $E_2(z)$ is an almost holomorphic modular form of weight $2$ and level $1$, so the automorphic representation generated by it is not irreducible. For more details (and my thought process), read below.
Consider the Maass raising operator in weight $0$,
$$ R:=y\left(i\frac{\partial}{\partial x}+\frac{\partial}{\partial y}\right).$$
Let $(m,n)\in\mathbb{Z}^2$ be a nonzero pair of integers. Then a small calculation gives that, for $z=x+iy$ and $s\in\mathbb{C}$,
$$ R\left(\frac{y^s}{|mz+n|^{2s}}\right) =\frac{sy^s}{(mz+n)^2|mz+n|^{2s-2}}.$$
Now let us introduce the usual weight $0$ level $1$ (nonholomorphic) Eisenstein series
$$ E(z,s):=\sum_{\substack{m, n \in \mathbb{Z} \\ (m, n) \ne (0,0)}} \frac{\operatorname{Im}(z)^s}{|mz + n|^{2s}},\qquad \operatorname{Re}(s)>1$$
then we see that
$$ R\,E(z,s+1) = (s+1)\,yE_2(z,s),\qquad \operatorname{Re}(s)>0.\tag{1}$$
On the right hand side, $yE_2(z,s)$ is the canonical weight $2$ level $1$ (nonholomorphic) Eisenstein series, the one which transforms as a weight $2$ Maass form. It is worthwhile to recall here that weight $k$ holomorphic forms embed into the weight $k$ Maass spectrum by multiplying each weight $k$ holomorphic form by $y^{k/2}$. In our case $k=2$, which explains why we multiply by $y$.
So your Eisenstein series, after inserting the factor $y$ to make it into a canonical weight $2$ form, and also inserting the scaling factor $s+1$, equals the Maass raising shift of $E(z,s+1)$. It belongs to the same automorphic representation as $E(z,s+1)$, hence it has the same Langlands parameters as $E(z,s+1)$ at every place. In particular, the archimedean Langlands parameters are
$$ (s+1)-\frac{1}{2}=s+\frac{1}{2}\qquad\text{and}\qquad \frac{1}{2}-(s+1)=-s-\frac{1}{2}.$$
Added and revised. Well, we still need to specify all this to $s=0$, but in this case the above argument breaks down, because $E(z,s+1)$ has a pole at $s=0$. So we need to be more careful. Let us use the results and notation of Section 4 of Duke-Friedlander-Iwaniec: The subconvexity problem for Artin L-functions (Invent. Math. 149 (2002), 489-577). Then for $\operatorname{Re}(s)>1$ we have the Fourier decomposition
\begin{align*} \frac{1}{2}E(z,s)&=\ \zeta(2s)y^s+\pi^{2s-1}\frac{\Gamma(1-s)}{\Gamma(s)}\zeta(2-2s)y^{1-s}\\&+\ \frac{\pi^s}{\Gamma(s)}\sum_{n=1}^\infty\frac{\sigma_{2s-1}(n)}{n^s}\bigl\{f_0^+(nz,s)+f_0^-(nz,s)\bigr\}.\end{align*}
Let us replace $s$ by $s+1$ here, and then apply the raising operator $R$ along with $(1)$. Then for $\operatorname{Re}(s)>0$ we obtain the Fourier decomposition
\begin{align*} \frac{1}{2}E_2(z,s)&=\ \zeta(2s+2)y^s+\pi^{2s+1}\frac{\Gamma(1-s)}{\Gamma(2+s)}\zeta(-2s)y^{-s-1}\\&-\ \frac{\pi^{s+1}}{y\Gamma(2+s)}\sum_{n=1}^\infty\frac{\sigma_{2s+1}(n)}{n^{s+1}}\bigl\{f_2^+(nz,s+1)+s(s+1)f_2^-(nz,s+1)\bigr\}.\end{align*}
The right hand side is indeed holomorphic at $s=0$, and at this value it specifies to
\begin{align} \frac{1}{2}E_2(z)&=\ \frac{\pi^2}{6}-\frac{\pi}{2y}- \frac{\pi}{y}\sum_{n=1}^\infty\frac{\sigma_1(n)}{n}f_2^+(nz,1)\\
&=\ \frac{\pi^2}{6}-\frac{\pi}{2y}-4\pi^2\sum_{n=1}^\infty\sigma_1(n)e(nz).\tag{2}\end{align}
It is clear now that the $L$-function of $E_2(z)$ is $\zeta(s-1/2)\zeta(s+1/2)$, and $E_2(z)$ should belong to the holomorphic discrete series of weight $2$ even though its constant term is not holomorphic. I think this paradox arises from the fact that $E_2(z)$ is not a true automorphic form. (More precisely, $E_2(z)$ is not a vector from an irreducible automorphic representation, see the Added 3 section below.)
Added 2. Indeed, $E_2(z)$ is an almost holomorphic modular form of weight $2$ and level $1$: it equals $2G_2^*(z)$ in the notation of Section 2.3 of Bruinier-v.d.Geer-Harder-Zagier's book "The 1-2-3 of modular forms". In particular, (19) and (21) in this book reveal that $E_2(z)$ transforms precisely like a holomorphic modular form of weight $2$ and level $1$, even though it is not holomorphic. Of course, the same also follows from the fact that $E_2(z)=\lim_{s->0+}E_2(z,s)$, where $yE_2(z,s)$ for $\operatorname{Re}(s)>0$ transforms precisely like a Maass form of weight $2$ and level $1$. One can read more about almost holomorphic modular forms in Section 5.3 of the book.
Added 3. We can learn more by applying the Maass lowering operator in weight $2$,
$$L:=1+y\left(i\frac{\partial}{\partial x}-\frac{\partial}{\partial y}\right).$$
Using the Fourier decomposition (2), we can see directly that
$$ L(yE_2(z))=-\pi, $$
which harmonizes with the facts that $ L(yE_2(z,s)) = -s E(z,s+1)$ for $\operatorname{Re}(s)>0$ and $\operatorname{res}_{s=0}E(z,s+1)=\pi$. So we see that the automorphic representation generated by the weight $2$ vector $yE_2(z)$ is reducible: it contains the trivial representation $\mathbb{C}$ as a subrepresentation, while its quotient by $\mathbb{C}$ is irreducible and belongs to the holomorphic discrete series of weight $2$. See also Theorem 2.5.2 in Bump: Automorphic forms and representations, specified to $k=2$ and $\lambda=0$, which helped me understand what is going on.
