$su(1,1) \cong su(2)$? The three generators of $su(2)$ satisfy the commutation relations
$$ [J_0 , J_\pm] = J_\pm , \quad [J_+, J_- ] = +2J_0  .$$ 
The three generators of $su(1,1)$ satisfy the commutation relations
$$ [K_0 , K_\pm] = K_\pm , \quad [K_+, K_- ] = -2K_0  .$$
Now, let us define 
$$ K_0 = J_0, \; K_+ = J_+,\; K_- = - J_-.  $$
It is apparent that so defined $K$'s satisfy the $su(1,1)$ algebra! Does this mean that $su(1,1)$ is actually equivalent to $su(2)$? 
Where is the argument wrong? 
 A: The ladder operators do belong to the real Lie algebras$^1$
$$\begin{align}
su(1,1)~:=~&\{m\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid m^{\dagger}\sigma_3=-\sigma_3m,~ {\rm tr}(m)=0\}\cr
~=~&{\rm span}_{\mathbb{R}}\{ \sigma_1, \sigma_2, i\sigma_3 \}\cr
~\cong~sl(2,\mathbb{R})
~:=~&\{m\in {\rm Mat}_{2\times 2}(\mathbb{R}) \mid  {\rm tr}(m)=0\}\cr
~=~&{\rm span}_{\mathbb{R}}\{  \sigma_1, i\sigma_2,  \sigma_3 \}\cr
~=~&{\rm span}_{\mathbb{R}}\{  \sigma_+, \sigma_-,  \sigma_3 \}\cr
~\cong~ so(2,1)~:=~&\{m\in {\rm Mat}_{3\times 3}(\mathbb{R}) \mid m^{t}\eta =- \eta m\},
\end{align}$$
$$ \sigma_{\pm}~:=~\frac{\sigma_1\pm i \sigma_2}{2}, \qquad \eta~=~{\rm diag}(1,1,-1), $$
but they do not  belong to the real Lie algebras
$$\begin{align}
su(2)~:=~&\{m\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid m^{\dagger}=-m,~ {\rm tr}(m)=0\}\cr
~=~&{\rm span}_{\mathbb{R}}\{ i\sigma_1, i\sigma_2, i\sigma_3 \}  \cr
~\cong~ so(3)~:=~&\{m\in {\rm Mat}_{3\times 3}(\mathbb{R}) \mid m^{t}=-m\}.
\end{align}$$
All the above 5 real Lie algebras have complexifications isomorphic to $$\begin{align}sl(2,\mathbb{C})~:=~&\{m\in {\rm Mat}_{2\times 2}(\mathbb{C}) \mid  {\rm tr}(m)=0\}\cr
~=~&{\rm span}_{\mathbb{C}}\{ \sigma_1, \sigma_2, \sigma_3 \}.\end{align}$$
--
$^1$ Here we follow the mathematical definition of a real Lie algebra. Be aware that in much of the physics literature, the definition of a real Lie algebra is multiplied with a conventional extra factor of the imaginary unit $i$, cf. footnote 1 in my Phys.SE answer here.
A: You may indeed identify the generators in the way you did. However, the Lie algebras and Lie groups are different because – as quickly said by Qmechanic – you must use different reality conditions for the coefficients.
A general matrix in the $SU(2)$ group is written as
$$ M = \exp[ i(   \alpha J_+ + \bar\alpha J_- + \gamma J_0 )] $$
where $\alpha\in {\mathbb C}$ and $\gamma\in {\mathbb R}$ while the general matrix in $SU(1,1)$ is given by 
$$ M' = \exp [ i( \alpha_+ J_+ + \alpha_- J_- + \beta J_0   ] $$
where $\alpha_+,\alpha_-,\beta\in {\mathbb R}$ are three different real numbers.
To summarize, for $SU(2)$, the coefficients in front of $J_\pm$ are complex numbers conjugate to each other, while for $SU(1,1)$, they are two independent real numbers. (And I apologize that I am not sure whether the $i$ should be omitted in the exponent of $SU(1,1)$ only according to your convention. Probably.)
If you allow all three coefficients in front of $J_\pm,J_0$ to be three independent complex numbers, you will obtain the complexification of the group. And as Qmechanic also wrote, the complexification of both $SU(2)$ and $SU(1,1)$ is indeed the same, namely $SL(2,{\mathbb C})$.
