First: The definition of Lie groups and a Lie algebras can vary depending on who you ask.
Given a (matrix) Lie group $G$ (so $G\subseteq GL_n(\mathbb{C})$), the Lie algebra of $G$ is the set
$$
\mathfrak{g} = \{X \in M_n(\mathbb{C}) \mid e^{tX} \in G \; \forall\; t\in \mathbb{R}\}.
$$
This is clearly then a real vector space. But it isn't necessarily a complex vector space. We always get a real Lie algebra. We only get a complex Lie algebra if $iX\in \mathfrak{g}$ for all $X\in \mathfrak{g}$. So just because the entries are complex, doesn't mean that the Lie algebra is complex.
Specifically considering $\mathfrak{sl}_2(\mathbb{C})$ you get the set of $2\times 2$ matrices with complex entries of trace zero. Again this is automatically a real vector space, but we can try to check if it is a complex vector space. We check that
$$
i\pmatrix{a & b \\ c & -a} = \pmatrix{ia & ib \\ ic & -ia}.
$$
This again has trace zero, so indeed $\mathfrak{sl}_2(\mathbb{C})$ is a complex Lie algebra. We usually then call the Lie group complex if the Lie algebra turns out to be a complex Lie algebra. Other complex Lie groups are $GL_n(\mathbb{C}), SL_n(\mathbb{C})$, $SO_n(\mathbb{C})$, and $Sp_n(\mathbb{C})$.
Another example: Consider the Lie group $SU(n)$ of all $n\times n$ unitary matrices (with entries from $\mathbb{C}$) with determinant $1$. In this case you can find that the Lie algebra $\mathfrak{su}(n)$ is the space of all $n\times n$ complex matrices $X$ where $X^* = -X$ ($*$ being complex conjugate transposed) and with trace $0$. This is not a complex Lie algebra, but only a real Lie algebra. So $SU(n)$ is not a complex Lie group.
Hopefully I didn't say anything wrong.
Very short answer: You have to be very precise about which base field, $\mathbb R$ or $\mathbb C$, you are considering in each case.
Over $\mathbb C$, there is the Lie group $SL_n(\mathbb C)$ and its Lie algebra $\mathfrak{sl}_n(\mathbb C)$, and every Cartan subalgebra of this will have roots which form a system of type $A_{n-1}$. There is extensive literature on this.
Over $\mathbb R$ however, one can e.g. look at the Lie groups $SL_n(\mathbb R)$, which have Lie algebra $\mathfrak{sl}_n(\mathbb R)$, but also at the Lie groups $SU(n)$ and their Lie algebras $\mathfrak{su}_n$ -- note that elements of these are often written as certain matrices with complex entries, but they are not complex Lie groups resp. algebras, but real ones. Notice in particular that $\mathfrak{su}_n$, which indeed can be identified with the traceless skew-hermitian $n\times n$-matrices, is not a vector space over $\mathbb C$, but over $\mathbb R$ (of dimension $n^2-1$).
Now it turns out that the non-isomorphic real Lie groups $SL_n(\mathbb R)$ and $SU_n$ both have complexification (isomorphic to) $SL_n(\mathbb C)$. They are so-called real forms of $SL_n(\mathbb C)$. Likewise but even simpler to see -- on the Lie algebra level, complexification is just done by tensoring with $\mathbb C$ -- both $\mathbb C \otimes_{\mathbb R}\mathfrak{su}_n$ and $\mathbb C \otimes_{\mathbb R}\mathfrak{sl}_n(\mathbb R)$ are isomorphic to $\mathfrak{sl}_n(\mathbb C)$, i.e. both $\mathfrak{su}_n$ and $\mathfrak{sl}_n(\mathbb R)$ are real forms of $\mathfrak{sl}_n(\mathbb C)$.
For $n=2$, $SL_2(\mathbb R)$ and $SU_2$ are (up to isomorphism) the only real forms of $SL_2(\mathbb C)$. For higher $n$ though, and for other classes of Lie groups / algebras, there are usually more real forms. The last example here is a real form of $\mathfrak{sl}_3(\mathbb C)$, called $\mathfrak{su}_{1,2}$, which is neither isomorphic to $\mathfrak{sl}_3(\mathbb R)$ nor to $\mathfrak{su}_3$.
It's quite common in the literature when speaking of root systems, what is meant is actually the root system of the complexification. In that terminology, both $SL_n(\mathbb R)$ and $SU_n$ (or their Lie algebras) have root system $A_{n-1}$. However, there is also the notion of relative or restricted or real or $k$-rational (here for $k=\mathbb R$) root systems; in this case, the relative root system of $SL_n(\mathbb R)$ would still be $A_{n-1}$, whereas the relative root system of $SU_n$ is empty (which is always the case for compact semisimple groups). More on those "relative roots" e.g. here, where I tried to compute all examples of real forms where that restricted root system is of type $BC$ (something that can never happen for complex Lie groups / algebras).
One further thing to note: By a fantastic coincidence (?), for each complex simple Lie algebra, there is up to iso exactly one real form which is compact (e.g. above, $\mathfrak{su}_n$ is the compact real form of $\mathfrak{sl}_n(\mathbb C)$). Also, there is always exactly one so-called "split" real form, whose restricted roots are just the same as the roots of the complexified version (e.g. above $\mathfrak{sl}_n(\mathbb R)$ is the split real form of $\mathfrak{sl}_n(\mathbb C)$). In a way, these two are extreme cases on opposite ends of a spectrum. As noted above, in general there are many more cases "between" them. They are classified by so-called "Satake diagrams", which are like an upgrade of the Dynkin diagrams: the underlying Dynkin diagram of a Satake diagram tells us of what type ($A_n, B_n, C_n, ..., G_2$) the complexification is, and the extra ornaments that make it a Satake diagram (black vs. white nodes, and arrows) encode which real form of that complex type we have. See further references and examples here or here.
Added: It's maybe worthwhile to note that beyond everything mentioned above, the (Lie group / Lie algebra)-correspondence is also not one-to-one, over any ground field. Rather, for one given semisimple Lie algebra there is a lattice of connected groups that "sits" over it, with an adjoint (centreless) one at the bottom and a simply connected one on top. E.g. over $\mathbb C$,
$PSL_2(\mathbb C)$ (adjoint) and $SL_2(\mathbb C)$ (simply connected) share the Lie algebra $\mathfrak{sl}_2(\mathbb C)$;
whereas over $\mathbb R$,
$PSL_2(\mathbb R)$ (adjoint), $SL_2(\mathbb R)$, $Mp_2(\mathbb R)$ (the metaplectic group), ... , $\overline {SL_2(\mathbb R)}$ (the simply connected universal cover of $SL_2(\mathbb R)$), with the "..." being infinitely more in-between, all share the Lie algebra $\mathfrak{sl}_2(\mathbb R)$ (compare last three sentences here);
whereas the compact real one has only two manifestations again:
$PSU_2$ (adjoint, and happens to be $\simeq SO_3(\mathbb R)$) and $SU_2$ (simply connected) share the Lie algebra $\mathfrak{su}_2$.
If one allows even disconnected groups, then there's infinitely many more groups sitting over each Lie algebra, but that's basically stuff like
$SL_2(\mathbb C) \times$ (your favourite finite group) still has Lie algebra $\mathfrak{sl}_2(\mathbb C)$.
Best Answer
There is a purely geometrical explanation. First, let $n=2$, and consider $\mathrm{SL}(2,\mathbb{R})$.
Let $X : [0,1) \to \mathrm{SL}(2,\mathbb{R})$ be a smooth path, with $X(0) =E$, where $E$ is the identity.
$$X(t) = \left[\begin{array}{cc} a(t) & b(t) \\ c(t) & d(t) \end{array}\right]$$ where $\det[X(t)] = (ad-bc)(t) = 1$ for all $t$. The derivative $X'(t)$, as $t$ tends to zero, gives a tangent vector to $\mathrm{SL}(2,\mathbb{R})$ at $E$. We have $$X'(t) = \left[\begin{array}{cc} a'(t) & b'(t) \\ c'(t) & d'(t) \end{array}\right]$$
Since $ad-bc \equiv 1$ we can differentiate to give $$a'd+ad'-b'c-bc'\equiv 0$$ Since $X(0)=E$ we know that $a(0)=1$, $b(0)=0$, $c(0)=0$ and $d(0)=1$. Hence: $$\begin{eqnarray*} a'd+ad'-b'c-bc' &\equiv& 0 \\ \\ a'(0)\,d(0)+a(0)\,d'(0)-b'(0)\,c(0)-b(0)\,c'(0) &=& 0 \\ \\ a'(0)+d'(0) &=& 0 \end{eqnarray*}$$ This tells us that $X'(0)$ must be a traceless matrix.
General Dimension
Jacobi's Formula tells us that
$$\det(X)' = \mathrm{tr}\! \left[ \mathrm{adj}(X)X' \right] $$
Since $\det(X) \equiv 1$ we have $\det(X)'\equiv 0$. When $t=0$, $X=E$ and so $\mathrm{adj}(X) = E$. Hence $$\mathrm{tr}\!\left[X'(0)\right]=0$$