The goal of my answer is only to append further definitions and references to Andrei Smolensky’s accepted answer.

The OP’s first question asks whether $SL_n(R)$, for $R$ a principal ideal domain (PID), is generated by the images of the fundamental embeddings of $SL_2(R)$. Andrei Smolensky suggests that it can be verified by inspecting the proof of the Smith Normal Form theorem for PIDs. Indeed, every matrix used in the reduction is either an elementary matrix (i.e., a matrix that differs from the identity by an off-diagonal element) or a matrix coming from one of the fundamental embeddings. Hence the result follows from Aurel’s comment.

As indicated by Andrei Smolensky, this result can be generalized both to rings of stable range at most $2$ and to Hermit rings in the sense of Kaplansky, using the spirit of the Smith Normal Form theorem.

The *stable rank* of a ring $R$ is, if it exists, the least integer $n$ such that for every $(n + 1)$-row $(r_1, \dots, r_{n + 1}) \in R^{n + 1}$ verifying
$r_1 R + \dots + r_{n + 1}R = R$ we can find $(\lambda_1, \dots, \lambda_n) \in R^n$ such that $(r_1 + \lambda_1 r_{n + 1})R + \dots + (r_n + \lambda_n r_{n + 1})R =R$.

**Assertion.** If the stable rank of $R$ is at most $2$, then $SL_n(R)$ is generated by the images of the fundamental embeddings of $SL_2(R)$.

**Proof.** Let $A \in SL_n(R)$. Since the first column of $A$ generates $R$, we can find a product $E$ of elementary matrices such that the coefficients of index $(1, 1)$ and $(2, 1)$ of $A’ = EA$ generates $R$. Using the top left-hand corner embedding, we can turn these two coefficients respectively into $1$ and $0$. Using the coefficient of index $(1,1)$ as a pivot we can reduce to a matrix in $SL_{n - 1}(R)$ and conclude by induction.

As Dedekind domains’ Krull dimension is at most $1$, it follows from the Bass Cancellation Theorem that Dedekind domains have stable rank at most $2$. This provides us with examples of rings which fail simultaneously to be Bézout rings and generalized Euclidean rings (a *generalized Euclidean ring* is a ring $R$ such that $SL_n(R)$ is generated by elementary matrices for every $n \ge 2$.)

A Bézout domain has stable rank at most $2$, see this MO post. Therefore, the above assertion answers also OP’s second question.

A Bézout domain is also a commutative Hermit ring in the sense of Kaplansky [Theorem 3.2, 1], i.e., a ring such that, for every $(r,s) \in R^2$ there is $A \in SL_2(R)$ and $d \in R$ such that $(r, s)A = (d, 0)$.

Hence the same answer can be inferred from

**Assertion.** If $R$ is a Hermit ring in the sense of Kaplansky, then $SL_n(R)$ is generated by the images of the fundamental embeddings of $SL_2(R)$.

**Proof.** Let $A \in SL_n(R)$. Using the top left-hand corner embedding, we can cancel the coefficient of index $(1, 2)$. Using a row permutation followed by a matrix from the latter embedding, we can cancel all coefficients in the first column except the coefficient of index $(1, 1)$. Iterating over columns, we can thus turn $A$ into a trigonal matrix $T$ whose diagonal coefficients are necessarily invertible.
Further elementary row transformations can then be used to turn $T$ into a diagonal matrix $D$. By Whitehead's lemma, $D$ can be turned into the identity matrix using elementary row transformations as well.

The class of commutative Hermit rings encompasses the class of commutative principal ideal rings [2], Bézout rings with finitely many minimal prime ideals [3, Theorem 2.2] and Bézout rings whose zero divisors lie in the nil radical [1, Theorem 3.2].

[1] "Elementary divisors and modules", I. Kaplansky, 1949 (MR0031470).

[2] "On the structure of principal ideal rings", T. W. Hungerford, 1968 (MR0227159).

[3] "Elementary divisors and finitely presented modules", M. Larsen, J. Lewis and T. Shores, 1974 (MR0335499).