This is an expansion of my comment.
The Smith normal form is a normal form of a matrix with entries in any given PID (but this probably works for non-domains and for Bezout rings in general). It goes as follows: given an $m\times n$ matrix $A$, there exist invertible $m\times m$ and $n\times n$ matrices $B$ and $C$ such that $BAC=\operatorname{diag}(a_1,\ldots,a_r,0,\ldots,0)$. Moreover, the entries $a_i$ satisfy $a_i\mid a_{i+1}$ and are unique up to the multiplication by a unit. This decribes the representatives of $GL(m,R) \backslash M(m,n,R) / GL(n,R)$. From here you can obtain the decription for $SL(m,R) \backslash M(m,n,R) / SL(n,R)$ — no independent multiplication by a unit anymore, so the difference is the same as between $K_1$ and $SK_1$.
When $R$ is a Euclidean ring, one has an algorith for computing the Smith normal form. This works for any PID, in fact, but the resulting matrices $B$ and $C$ are not necessary elementary in this case. For a Euclidean ring the algorithm gives you the chain of elementary transfromation for obtaining the SNF.
As you mentioned, $E(n,R)=SL(n,R)$ when $R$ is Euclidean, but this is not the case for a PID. This equality fails, for example, for the ring $S^{-1}\mathbb{Z}[x]$, where $S$ is the multiplicative system generated by all cyclotomic polynomials. This is a result of
D. R. Grayson — $SK_1$ of an interesting principal ideal domain.
Here are two other links with examples:
The complete answer to your question would contain the computation of $K_1$ for an arbitrary PID, so basically there are no chances. Howerer, it is easy to show that using the elementary transformation you can almost get the SNF. Namely, the usual procedure for Euclidian rings will fail in the setting of PID only in the last step, when you are facing the problem of simplifying a $2\times2$ matrix. So each coset contains a block diagonal matrix with only one block of size >1 (which is of size 2). The determinants of the blocks satisfy the same relation as $a_i$'s.