The question is somewhat ambiguous, it’s not clear whether the Archimedean classes are meant to be additive or multiplicative. I will assume the former, i.e., equivalence classes of the relation $$a\sim b\Leftrightarrow\exists n,m\in\omega\smallsetminus\{0\}\,(na\le b\le ma).$$ b\Leftrightarrow\mathrm{sign}(a)=\mathrm{sign}(b)\land\exists n\in\omega\smallsetminus\{0\}\,(n^{-1}|a|\le|b|\le n|a|).$$ First, since real-closed fields (rcf) have elimination of quantifiers, any embedding between them is automatically elementary. Thus the question is whether every rcf $R$ has a proper rcf extension $S$ with the same Archimedean classes (i.e., every $s\in S$ is $\sim$ to some $r\in R$).
As Andreas noted above, this property does not hold in general, and in particular, $\mathbb R$ has no proper Archimedean extension. On the other hand, it holds for many other real-closed fields: for example, any Archimedean rcf different from $\mathbb R$ has a proper Archimedean rcf extension (namely, $\mathbb R$). I think the following characterization holds:
Proposition: If $R$ is a rcf, the following are equivalent:
$R$ has a proper rcf extension with the same Archimedean classes.
There is a Dedekind cut $\langle A,B\rangle$ on the interval $(0,1)_R$ such that $$\tag{$*$}\forall a\in A\,\exists b\in B\,\frac{a+b}2\in A\qquad\text{and}\qquad\forall b\in B\,\exists a\in A\,\frac{a+b}2\in B.$$
On the one hand, let $S\supseteq R$ be a rcf with the same Archimedean classes and $x\in S\smallsetminus R$. We can assume $x>1$. There exists $c\in R$ such that $c\sim x$; WLOG $c< x< 2c$. Then $0< x/c-1< 1$, and the Dedekind cut on $R$ determined by $x/c-1$ is easily seen to satisfy $(*)$.
On the other hand, assume the cut $\langle A,B\rangle$ is given. We define an ordering on the rational function field $F=R(x)$ as follows. Using the fact that every nonzero polynomial is a product of linear polynomials and polynomials of the form $(x-a)^2+b$, where $b>0$, we see that for every $f(x)/g(x)\in F$, there are $a\in A$ and $b\in B$ such that $f$ and $g$ have constant sign on $(a,b)_R$; we define the sign of $f(x)/g(x)$ to be the sign it assumes on $(a,b)_R$. This makes $F$ an ordered field. Let $S$ be its real closure. For a given $\alpha\in S$, there exists $c\in R$ such that $\alpha\sim c$ whenever:
$\alpha=x-a$, $a\in R$. This follows from $(*)$.
$\alpha=(x-a)^2+b$, $a,b\in R$, $b>0$. This follows from 1: if $u\sim u'$, $v\sim v'$, and $u,v>0$, then $u+v\sim u'+v'$.
$\alpha\in F$. Every such $\alpha$ is a product of an element of $R$ and elements of the form 1 or 2 or their inverses, and $u\sim u'$ and $v\sim v'$ imply $uv\sim u'v'$ and $u/v\sim u'/v'$.
$\alpha\in S$ is such that $\alpha^k\in F$ for some integer $k>0$. We have $\alpha^k\sim c$ by 3, hence $\alpha\sim\sqrt[k]c\in R$.
$\alpha\in S$. We have $\sum_{i\le d}u_i\alpha^i=0$ for some $u_i\in F$, $u_d\ne0$. Let $i$ be such that the Archimedean class of $u_i\alpha^i$ is maximal. Since the sum above is $0$, there exists $j\ne i$ such that $u_j\alpha^j\sim-u_i\alpha^i$. Then $\alpha^{j-i}\sim-u_i/u_j$, hence $\alpha\sim c$ for some $c\in R$ by 4.
Thus $S$ is a proper rcf extensions of $R$ with the same Archimedean classes as $R$.
