A pair of fields like this exists.
As noted in the question, one may assume that $E$ is a field of characteristic zero. Therefore $E^+$ is an infinite, torsion-free, divisible, abelian group; i.e. an infinite rational vector space. Moreover, any infinite rational vector space is isomorphic to the additive group of some field of characteristic zero. Thus the question reduces to this: is there an infinite field $F$ whose multiplicative group of units is a rational vector space?
Such a field is constructed in
Adler, Allan
On the multiplicative semigroups of rings.
Comm. Algebra 6 (1978), no. 17, 1751-1753.
Here is the first line of George Bergman's Math Review of this paper:
``Let $\mathbb Q$ denote the additive group of the rational numbers. It is shown that there exists a field whose multiplicative group is isomorphic to
$\mathbb Q^{\aleph_0}$, but none whose multiplicative group is isomorphic to $\mathbb Q$.''
EDIT: 8/21/20
In response to the August 18 question of mr_e_man below, I asked my library to get a scan of Adler's paper. Adler's construction of a field whose multiplicative group is isomorphic to the direct power $\mathbb Q^{\aleph_0}$ goes like this:
Let $\mathbb F_{2^n}$ be Galois field of order $2^n$. Let $P$ be the set of primes. The desired field can be taken to be any field $\mathbb K$ that is an ultraproduct of the form $\prod_{\mathcal U} \mathbb F_{2^p}$ over a nonprincipal ultrafilter $\mathcal U$ defined on $P$. Let's show that the multiplicative group of $\mathbb K$ is isomorphic to $\mathbb Q^{\aleph_0}$.
Up to isomorphism, $\mathbb Q^{\aleph_0}$ is the unique torsion-free divisible abelian group of cardinality continuum.
The multiplicative group $\mathbb K^{\times}$ is isomorphic to
the abelian group
$\prod_{\mathcal U} \mathbb F_{2^p}^{\times}$. The factors in this ultrproduct have sizes $2^2-1< 2^3-1 < 2^5-1 < \cdots$, so the sizes are a strictly increasing countable sequence. This is enough to prove that the ultraproduct has size continuum. (I am asserting that any nonprincipal ultraproduct of countably many finite sets of strictly increasing size has cardinality continuum.)
If $p\neq q$, then $\gcd(|\mathbb F_{2^p}^{\times}|,|\mathbb F_{2^q}^{\times}|)=\gcd(2^p-1,2^q-1)=1$. Thus, for any $m\leq \textrm{min}(2^p-1, 2^q-1)$, at most one of the groups $\mathbb F_{2^p}^{\times}$ has $m$-torsion. In fact, the map $x\mapsto x^m$ must be a bijection on all $\mathbb F_{2^p}^{\times}$ except possibly finitely many of them. In the ultraproduct $\prod_{\mathcal U} \mathbb F_{2^p}^{\times}=\mathbb K^{\times}$ the maps $x\mapsto x^m$, $m\in\mathbb Z^+$, must all be bijections. This is enough to show that $\mathbb K^{\times}$ is torsion-free and divisible.
$\newcommand{\N}{\mathbb N} \newcommand{\Z}{\mathbb Z} \newcommand{\Q}{\mathbb Q}$ Actually I think the claim to be false. A counter example should be given by $\Q^*$.
The short version is to observe that $\Q^*$ is isomorphic to $\Z/2\Z \times \bigoplus_{n \in \N} \Z$, then by the following chain of isomorphisms
$$\Z/2\Z \times \bigoplus_{n \in N} \Z \cong \Z/2\Z \times \Z \times \bigoplus_{n \in N} \Z \cong \Z \times \Z/2\Z \times \bigoplus_{n \in N} \Z \cong \Z \times \Q^*$$
it follows that $\Q^* \cong \Z \times \Q^*$.
To see the isomorphism between $\Q^*$ and $\Z/2\Z \times \bigoplus_{n \in \N} \Z$ consider the following map
$$
\begin{align*}
f \colon \Z/2\Z \times \bigoplus_{n \in \N} \Z & \longrightarrow \Q^* \\
f(s, \bar a) = (-1)^s \prod_{n \in \N} p_n^{a_n}
\end{align*}
$$
where $s \in \Z/2\Z$, $\bar a =(a_n) \in \bigoplus_n \Z$ and for each natural number $n$ the number $p_n$ is the $n$-th prime number in $\N$.
By a simple calculation one can prove that $f$ is indeed a group homomorphism:
$$
\begin{align*}
f((s,\bar a)+(t,\bar b)) &= f((s+t,\bar a + \bar n)) \\
&= (-1)^{s+t} \prod_n p_n^{a_n+b_n} \\
&= (-1)^s \prod_n p_n^{a_n} (-1)^t \prod_n p_n^{b_n} \\
&= f((s,\bar a)) f((t,\bar b))
\end{align*}
$$
The injectivity follows by observing that $f((s,\bar a)) = 1$ if and only if
$(-1)^s\prod_n p_n^{a_n} = 1$, which happens only if $s=0$ and for every $n$ we have $p_n^{a_n}=1$, that is if $a_n=0$.
Surjectivity follows observing that the image of $f$ contains all the integers which generate $\Q^*$.
I hope this helps.
Best Answer
Of course.
There trivial groups $\{0\}, +$ and $\{1\}, \times$ but we could have general cyclic groups with the same order.
Example in $\mathbb R$ then $<1,+> = \mathbb Z$ is isomorphic to $\{q^n|n\in \mathbb Z\}, \times$ (assuming $q>0; q\ne 1$).