I believe this could be related to an algebraic structure useful for studying set-theoretical solutions of the Yang-Baxter equation, the so-called braces. At least the funny distributive property appears in the theory of set-theoretical solutions of the Yang-Baxter equation.
Definition. An abelian group $(A,+)$ is a left brace if there is a multiplication $m\colon A\times A\to A$, $(a,b)\mapsto ab$, turning $A$ into a group such that
$$a(b+c)+a=ab+ac$$
holds for all $a,b,c\in A$.
Similarly one defines right braces.
The connection to the Yang-Baxter equation is as follows. First, for each left brace $A$
one has a canonical solution $(A,r_A)$ of the Yang-Baxter equation:
$$r_A\colon A\times A\to A\times A,\quad r_A(a,b)=(ab-a,(ab-a)^{-1}ab).$$
One also has the following.
Theorem. Let $X$ be a set, $$r\colon X\times X\to X\times X,\quad r(x,y)=(\sigma_x(y),\tau_y(x))$$ be a
non-degenerate involutive solution of the Yang–Baxter equation. (Non-degenerate means that $\sigma_x$ and $\tau_x$ are permutations for each $x\in X$, and involutive means that $r^2=\mathrm{id}_{X\times X}$.) Then there exists a unique left brace structure over the group
$$G=G(X,r)=\langle X:xy=\sigma_x(y)\tau_y(x)\rangle$$
such that its associated solution $r_G$ satisfies
$r_G(\iota\times \iota) = (\iota\times \iota)r$, where $\iota\colon X\to G(X,r)$ is the canonical map.
Furthermore, if $B$ is a left brace and $f\colon X\to B$ is a map such that $(f\times f)r=r_B(f\times f)$,
then there exists a unique brace homomorphism $\varphi\colon G(X,r)\to B$ such that $f = \varphi\iota$.
and $(\varphi\times\varphi)r_G = r_B(\varphi\times\varphi)$.
The theorem is implicit in the work of Rump where braces were discovered:
You can find some interesting examples of braces here:
See also this survey:
- Rump, Wolfgang. The brace of a classical group. Note Mat. 34 (2014), no. 1, 115--144. MR3291816
There is a non-commutative version of braces useful to study
non-involutive solutions, see this paper.
Added. Rump proved that braces on both sides are in bijective correspondence with radical rings. The correspondence is as follows. If $R$ is a non-zero radical ring (for all $x\in R$ there is $y\in R$ such that $x+y+xy=0$) then $R$ with $a\circ b=ab+a+b$ is a two-sided brace. Conversely, if $A$ is a two-sided brace, then $A$ with $a*b=ab-a-b$ is a radical ring.
Your question is equivalent to whether the category $\mathcal{E}$ of pairs $(V,f)$ consisting of a finitely generated free (right) $R$-module and an endomorphism $f$ of $V$ is a Krull-Schmidt category, i.e., an additive category where every object decomposes as a direct sum of finitely many indecomposable objects and the decomposition is unique up to isomorphism and reordering. (The category $\cal E$ is also equivalent to the category of right $R[t]$-modules which are f.g. free $R$-modules, and it is rarely Krull-Schmidt, but I won't use this point of view.)
Here is one possible counterexample, phrased using the category $\mathcal{E}$ of pairs $(V,f)$ above: Take $R$ to be a Dedekind domain admitting a non-free rank-$1$ projective module $L$ such that $L\oplus L\cong R\oplus R$ (equivalently, $L$ represents an element of order $2$ in the Picard group of $R$). Let $f_L : L^2\to L^2$ be defined by $f_L(x,y)=(0,x)$, and define $f_R:R^2\to R^2$ similarly. The isomorphism $L\oplus L\cong R\oplus R$ gives rise to an isomorphism
$$(L^2,f_L)\oplus (L^2,f_L) \cong (R^2,f_R)\oplus (R^2,f_R)$$
in $\cal E$.
One readily checks that ${\rm End}_{\cal E}(L^2,f_L)\cong R[\epsilon|\epsilon^2=0]$ and ${\rm End}_{\cal E}(R^2,f_R)\cong R[\epsilon|\epsilon^2=0]$, so the endomorphism rings of $(L^2,f_L)$ and $(R^2,f_R)$ contain no nontrivial idempotents. This means that these objects are indecomposable in $\cal E$. On the other hand, $(L^2,f_L)\ncong (R^2,f_R)$ because $\ker f_L\cong L\ncong R\cong \ker f_R$. Consequently, the monoid $(\cal E/\cong, \oplus)$ (which is isomorphic to $(J(R)/\sim, \oplus)$ in your question) is not a free abelian monoid (because $2x=2y$ implies $x=y$ in a free abelian monoid).
One the other hand, a sufficient (but not necessary) condition for the category $\cal E$ to be Krull-Schmidt is that the endomorphism ring of every object $(V,f)$, i.e., the centralizer of $f$ in ${\rm End}_R(V)\cong {\rm M}_n(R)$, is a semiperfect ring.
For example, if $R$ is commutative and aritinian as in Benjamin Steinberg's comment, then ${\rm End}_{\cal E}(V,f)$ will be an $R$-subalgebra of ${\rm End}_R(V)$, hence artinian, and in particular semiperfect.
The semiperfectness ${\rm End}_{\cal E}(V,f)$ for all f.g. free $V$ is actually true even if $R$ is non-commutative one-sided artinian, and even if $R$ is just semiprimary. This appears implicitly in a paper of mine (page 20 & Thm. 8.3(iii) & Remark 2.9).
When $R$ is commutative noetherian and local, one can use a theorem of Azumaya (Theorem 22) and a little work to show that this property is equivalent to $R$ being henselian.
Best Answer
There do exist pairs of finite unital rings whose additive structures are isomorphic and whose multiplicative structures are isomorphic, yet the rings themselves are not isomorphic.
To see this, let $\mathbb F$ be a field and let $X = \{x_1,\ldots, x_n\}$ be a set of variables. The polynomial ring $\mathbb F[X]$ is graded by degree $$ \mathbb F[X] = H_0\oplus H_1\oplus H_2\oplus\cdots. $$ Let $Q(x_1,\ldots,x_n)$ be a quadratic form over $\mathbb F$. Let $$I = \mathbb F\cdot Q(X)\oplus H_3\oplus H_4\oplus\cdots$$ be the ideal generated by $Q(X)$ and the homogeneous components of degree at least $3$. Let $S_{\mathbb F,Q}$ denote the $\mathbb F$-algebra $\mathbb F[X]/I$. It is a commutative, local ring, which encodes properties of the quadratic form $Q$.
Two quadratic forms $Q_1$ and $Q_2$ are equivalent if they differ by an invertible linear change of variables.
Claim. Let $\mathbb F$ be a finite field of odd characteristic $p$. Let $Q_1(x_1,\ldots,x_n)$ and $Q_2(x_1,\ldots,x_n)$ be nonzero quadratic forms over $\mathbb F$.
$S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ have isomorphic $\mathbb F$-space structures.
If $n>4$ and $Q_1$ and $Q_2$ are nondegenerate, then $S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ have isomorphic multiplicative monoids.
$S_{\mathbb F,Q_1}\not\cong S_{\mathbb F,Q_2}$ as $\mathbb F$-algebras, unless $Q_1$ is equivalent to a nonzero scalar multiple of $Q_2$.
Proof. Exercise! \\
So let $\mathbb F = \mathbb F_3$ be the $3$-element field. It is known that over a finite field of odd characteristic the quadratic forms are classified by the dimension and by the determinant of the form modulo squares. The determinant of $$ Q(x_1,\ldots,x_n)=a_1x_1^2+a_2x_2^2+\cdots+a_nx_n^2 $$ is $a_1\cdots a_n$. If $\alpha\in \mathbb F_3^{\times}=\{\pm 1\}$, then $\alpha\cdot Q$ has determinant $\alpha^n a_1\cdots a_n=(\pm 1)^n a_1\cdots a_n$. If $n$ is even, then the determinants of $Q$ and $\alpha\cdot Q$ will be equal, so $Q$ will be equivalent to $\alpha\cdot Q$ for every $\alpha\in \mathbb F_3^{\times}$. This implies that, when working over $\mathbb F_3$ in an even dimension, if $Q_1$ is not equivalent to $Q_2$, $Q_1$ will also not be equivalent to any nonzero scalar multiple of $Q_2$. In particular, no scalar multiple of $$ Q_1 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2, $$ is equivalent to $$ Q_2 = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 - x_6^2 $$ over $\mathbb F_3$. For these forms we have that $S_{\mathbb F,Q_1}$ and $S_{\mathbb F,Q_2}$ are nonisomorphic finite unital rings with isomorphic additive and multiplicative structures. (These rings have size $3^{27}$.)
Minor side comment 1: If you allow nonunital rings, there is a pair of nonisomorphic $8$-element rings whose additive and multiplicative structures are isomorphic.
Minor side comment 2: The solution to the exercise above (that is, the proof of the Claim) can be found here.