One is tempted (as I was originally), to argue as follows: since $M\cong M\oplus M\cong M\times M$, we have
$$R = \mathrm{Hom}(M,M) \cong \mathrm{Hom}(M,M\times M) \cong \mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M) = R\times R,$$
and stop there. The problem with the argument is that in some instances we are dealing with these objects as $\mathbb{Z}$-modules rather than as $R$-modules (specifically the isomorphisms are isomorphism from $\mathrm{Hom}(M,M)$ to $\mathrm{Hom}(M,M\times M)$ is, right now, just an isomorphism as $\mathbb{Z}$-modules), so some care needs to be exercised to make the argument actually work.
First, note that $M\cong M\times M$ as $\mathbb{Z}$-modules.
Define a homomorphism $\varphi\colon M\to M\times M$ by maping $(m_i)$ to $\bigl((m_{2i-1}), (m_{2i})\bigr)$. That is, $(m_1,m_2,m_3,\ldots)$ maps to $\bigl( (m_1,m_3,m_5,\ldots),(m_2,m_4,m_6,\ldots)\bigr)$.
The elements of $R$ can be thought of as infinite "column-finite matrices"; that is, each endomorphism $M\to M$ corresponds to a family of functions $(\mathbf{f}_i)_{i\in\omega}$, with $\mathbf{f}_i\colon\mathbb{Z}\to M$; hence $\mathbf{f}_i = \sum_{j\in\omega}f_{ji}e_j$, where $e_j$ is the element of $M$ that has $1$ in the $j$th coordinate and $0$s elsewhere, $f_{ji}\in\mathbb{Z}$, and $f_{ji}=0$ for almost all $j$.
If we take an element of $\mathbf{f}=(f_{ij})$ of $R$, and compose it with the isomorphism $M\to M\times M$, we obtain a homomorphism $M\to M\times M$, given by $\bigl( (f_{i,2j-1}), (f_{i,2j})\bigr)$. So the map $\mathrm{Hom}(M,M)\to\mathrm{Hom}(M,M)\times \mathrm{Hom}(M,M)$ maps $(f_{ij})$ to $\bigl( (f_{i,2j-1}),(f_{i,2j})\bigr)$.
If we compose maps $\mathbf{f}=(f_{ij})$ and $\mathbf{g}=(g_{ij})$, we get the map
$(h_{ij})$, where
$$h_{ij} = \sum_{k=1}^{\infty}g_{ik}f_{kj}.$$
Since $f_{kj}=0$ for almost all $k$, the sum is finite and $h_{ij}$ makes sense.
This gives the action of $R$ on $\mathrm{Hom}(M,M)$. The action of $R$ on $\mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M)$ is then given by
$$
\mathbf{g}\bigl((f_{ij}),(f'_{ij})\bigr) = \bigl( \mathbf{g}(f_{ij}), \mathbf{g}(f'_{ij})\bigr)
= \bigl( (h_{ij}), (h'_{ij})\bigr)$$
where
$$h_{ij} = \sum_{k=1}^{\infty}g_{ik}f_{kj},\quad\text{and}\quad h'_{ij}=\sum_{k=1}^{\infty}g_{ik}f'_{kj}.$$
To see that the map from $\mathrm{Hom}(M,M)$ to $\mathrm{Hom}(M,M\times M)$ respects the action of $R$ (that is, that we get an $R$-module homomorphism, not merely a $\mathbb{Z}$-module homomorphism), let $(f_{ij})\in\mathrm{Hom}(M,M)$ and $\mathbf{g}\in R$. The element $\mathbf{g}(f_{ij})$ is mapped to
$$\left( \Bigl(\sum_{k=1}^{\infty}g_{ik}f_{k,2j-1}\Bigr),\Bigl( \sum_{k=1}^{\infty}g_{ik}f_{k,2j}\Bigr)\right)$$
On the other hand, if we let $\mathbf{g}$ act on $\bigl( (f_{i,2j-1}), (f_{i,2j})\bigr)$, we get
$$\mathbf{g}\left( (f_{i,2j-1}), (f_{i,2j})\right) = \left(\Bigl( \sum_{k=1}^{\infty}g_{ik}f_{k,2j-1}\Bigr), \Bigl( \sum_{k=1}^{\infty}g_{ik}f_{k,2j}\Bigr)\right),$$
that is, the same as the image of $\mathbf{g}(f_{ij})$. So the homomorphism $\mathrm{Hom}(M,M)\to\mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M)$ given by $(f_{ij})\longmapsto ((f_{i,2k-1}),(f_{i,2k}))$ is actually a homomorphism as $R$-modules. Since the original map as $\mathbb{Z}$-modules was a bijection, so is this one, hence $\mathrm{Hom}(M,M)$ and $\mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M)$ are isomorphic not only as $\mathbb{Z}$-modules, but also as $R$-modules. So we obtain:
$$R = \mathrm{Hom}(M,M) \cong \mathrm{Hom}(M,M)\times\mathrm{Hom}(M,M) =R\times R,$$
with the isomorphism being an isomorphism of $R$-modules, as desired.
Here's a low-tech approach. Start with a simple case: suppose $a$ and $b$ are powers of some prime $p$, $a = p^i$ and $b = p^j$. Then $m = p^{\min(i, j)}$ and $n = p^{\max(i, j)}$. So $m$ and $n$ are the same as $a$ and $b$ (perhaps swapped), and there's clearly an isomorphism as required.
Now suppose there are two different primes involved, $p_0$ and $p_1$: $a = {p_0}^{i_0} {p_1}^{i_1}$, $b = {p_0}^{j_0} {p_1}^{j_1}$. Intuitively, because $p_0$ and $p_1$ are relatively prime, they operate independently. So $\Bbb{Z}/(a) \simeq \Bbb{Z}/({p_0}^{i_0}) \times \Bbb{Z}/({p_1}^{i_1})$ and $\Bbb{Z}/(b) \simeq \Bbb{Z}/({p_0}^{j_0}) \times \Bbb{Z}/({p_1}^{j_1})$. Also, $m = {p_0}^{\min(i_0, j_0)} {p_1}^{\min(i_1, j_1)}$ and $n = {p_0}^{\max(i_0, j_0)} {p_1}^{\max(i_1, j_1)}$, so $\Bbb{Z}/(m) \simeq \Bbb{Z}/({p_0}^{\min(i_0, j_0)}) \times \Bbb{Z}/({p_1}^{\min(i_1, j_1)})$ and $\Bbb{Z}/(n) \simeq \Bbb{Z}/({p_0}^{\max(i_0, j_0)}) \times \Bbb{Z}/({p_1}^{\max(i_1, j_1)})$. This shows that $\Bbb{Z}/(a) \times \Bbb{Z}/(b)$ and $\Bbb{Z}/(m) \times \Bbb{Z}/(n)$ are (up to isomorphism) just rearrangements of the same factors, so again we're done.
Finally we extend this argument to any number of primes $p_k$.
Best Answer
They are not isomorphic.
In what follows, $\bigoplus$ and $\prod$ will denote direct sums or products of copies of groups indexed by $\mathbb{N}$, and $e_{n}$ will denote the element of $\prod\mathbb{Z}$ whose $n$th coordinate is $1$, with all other coordinates zero.
I will show that there is no surjective homomorphism $\bigoplus\prod\mathbb{Z}\to\prod\bigoplus\mathbb{Z}$.
An abelian group $G$ is called slender if, for every homomorphism $\alpha:\prod\mathbb{Z}\to G$, $\alpha(e_{n})=0$ for all but finitely many $n$. The following properties of slender groups are standard (see, for example, Section 13.2 of Fuchs's 2015 book Abelian Groups).
$\mathbb{Z}$ is slender (this is just the well-known theorem of Specker).
(Infinite) direct sums of slender groups are slender. So in particular $\bigoplus\mathbb{Z}$ is slender.
The image of any homomorphism from $\prod\mathbb{Z}$ to a slender group is a finite rank free abelian group.
So for any homomorphism $\alpha:\prod\mathbb{Z}\to\bigoplus\mathbb{Z}$, there is some $n$ such that the image of $\alpha$ contains only elements of $\bigoplus\mathbb{Z}$ whose $m$th coordinates are zero for $m\geq n$. Call the least such $n$ the type of $\alpha$.
A homomorphism $\beta:\prod\mathbb{Z}\to\prod\bigoplus\mathbb{Z}$ consists of a sequence of homomorphisms $\beta_{i}:\prod\mathbb{Z}\to\bigoplus\mathbb{Z}$. If $\beta_{i}$ has type $n_{i}$, we'll say that $\beta$ has type $(n_{0},n_{1},n_{2},\dots)$.
A homomorphism $\gamma:\bigoplus\prod\mathbb{Z}\to\prod\bigoplus\mathbb{Z}$ consists of a sequence of homomorphisms $\gamma_{i}:\prod\mathbb{Z}\to\prod\bigoplus\mathbb{Z}$.
Suppose $\gamma_{i}$ has type $(n_{i,0},n_{i,1},n_{i,2},\dots)$. Then every element of the image of $\gamma$, considered as an $\mathbb{N}\times\mathbb{N}$ matrix with finitely supported rows, has the following property:
But one can easily construct elements of $\prod\bigoplus\mathbb{Z}$ that do not have this property, and so $\gamma$ cannot be surjective, and in particular cannot be an isomorphism. For example, take an $\mathbb{N}\times\mathbb{N}$ matrix where the $j$th row has a single nonzero entry, in the $N_{j}$th column, where $N_{j}=\max(n_{0,j},n_{1,j},\dots,n_{j,j})$.
In fact, something more general is true. Contained in Corollary 3 of
Zimmermann-Huisgen, Birge, On Fuchs’ problem 76, J. Reine Angew. Math. 309, 86-91 (1979). ZBL0408.20038.
is the fact that the groups $\bigoplus\mathbb{Z},\;\prod\mathbb{Z},\;\bigoplus\prod\mathbb{Z},\;\prod\bigoplus\mathbb{Z},\;\bigoplus\prod\bigoplus\mathbb{Z},\;\prod\bigoplus\prod\mathbb{Z},\dots$ are pairwise nonisomorphic.