L. N. Vaserstein's theorem [2] asserts that if $R$ is a Dedekind ring of arithmetic type with infinitely many units then $SL_2(R) = E_2(R)$ holds. In other words $R$ is a $GE_2$-ring in the sense of P. M. Cohn [1]. Here $E_2(R)$ denotes the subgroup of $SL_2(R)$ generated by the elementary matrices (also called transvections), i.e., the matrices of the form $\begin{pmatrix} 1 & r \\ 0 & 1\end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ r & 1\end{pmatrix}$ with $r \in R$.
Since the class of Dedekind rings of arithmetic type comprises the class of rings of integers of algebraic number fields, Vaserstein’s theorem answers OP’s question in the positive when the unit group is infinite.

But you may object, like B. Liehl [3], that there is a gap in Vaserstein’s proof. Fortunately A. Leutbecher [Section 2, 3] corrected this gap and B. Liehl [3] subsequently extended Vaserstein’s theorem to orders of arithmetic type with infinitely many units.

By Dedekind’s unit theorem, a ring of integers $R$ of an algebraic number field $K$ has finitely many units if and only if $K$ is either the field of rationals $\mathbb{Q}$ or an imaginary quadratic number field. Under this assumption P. M. Cohn proved that $SL_2(R) = E_2(R)$ if and only if $R$ is Euclidean with respect to complex modulus. Thus $\mathbb{Z}[\frac{1 + \sqrt{-19}}{2}]$ is a PID which is not a $GE_2$-ring. Remarkably, the subgroup $E_2(R)$ is non-normal in $SL_2(R)$ and has infinite index if $R$ is the ring of integers of an imaginary quadratic number field which is not Euclidean, see [Theorem 1.5, 6]. The proof of the latter exhibits infinitely many distinct coset representatives for the quotient $SL_2(R)/E_2(R)$, namely the so-called $S_{x, y}$ matrices, but not a transversal, a priori. The ring of integers $R$ of $\mathbb{Q}(\sqrt{-D})$ for $D = 5, 10$ and $14$ is scrutinized in [5] where a group presentation of $SL_2(R)$ is derived. I leave here the second part of OP's question, only partially answered.

These facts were already mentioned under the form of comments to OP’s question and are thoroughly discussed in B. Nica’s paper [6], which I warmly recommend. I put them in this answer because the question seems to be considered unsettled in a closely related MO post.

The following may help clear any doubt: If $R$ is an order in an algebraic number field $K$ which is not imaginary quadratic then $E_2(R)$ is a normal subgroup of $SL_2(R)$ of finite index. This is a result of [4], extracted under this form in [Theorem 1.6, 6]. The fact that the index of $E_2(R)$ in $SL_2(R)$ is actually $1$ if $R$ is moreover a maximal order, i.e., $R$ is the (full) ring of integers of $K$, may easily slip out of one’s mind.

[1] «On the structure of the $GL_2$ of a ring», P. M. Cohn, 1966.

[2] «On the group $SL_2$ over Dedekind rings of arithmetic type», L. N. Vaserstein, 1972.

[3] «Euklidischer Algorithmus und die Gruppe $GL_2$», A. Leutbecher, 1972.

[4] «On the group $SL_2$ over orders of arithmetic type», B. Liehl, 1981.
[5] "On the groups $SL_2(\mathbb{Z}[x])$ and $SL_2(k[x, y])$", F. Grunewald et al., 1994.

[6] «The Unreasonable Slightness of $E_2$ over Imaginary Quadratic Rings», B. Nica, 2013.

non-Euclideanin my comment. If you read Nica's survey even more carefully, you'd see that the imaginary quadratic fields for which there is elementary generation are exactly the Euclidean ones. – Sue Aug 30 '12 at 16:13