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There is in module theory the following (krull-shmidt) theorem:

If $(M_i)_{i\in I}$ and $(N_j)_{j\in J}$ are two families of simple module such that $\bigoplus\limits_{i\in I} M_i \simeq \bigoplus\limits_{j\in J} M_j$, then there exists a bijection $\sigma:I\rightarrow J$ such that $M_i \simeq N_{\sigma(i)}$.

Is there a similar kind of theorem for partially ordered sets? More precisely, is there a class of 'simple' partial orders such that if $(P_i)_{i\in I}$ and $(Q_j)_{j\in J}$ are 'simple' orders such that $\prod\limits_i P_i \simeq \prod\limits_j Q_j$, then there exists a bijection $\sigma:I \rightarrow J$ such that $P_i \simeq Q_{\sigma(i)}$.

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    How are you defining the order on $\prod\limits_i P_i$? Lexigraphically? – 2017-02-03
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    The standard ordering: $(x_i) \leq (y_i)$ iff $\forall i\in I(x_i \leq y_i)$ – 2017-02-03
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    As orders go, that is one of the less useful. I'd hardly call it "standard". – 2017-02-03
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    In the category of partially ordered sets and order-preserving maps, $\prod\limits_i P_i$ together with that ordering is the product of $(P_i)$ - this is what I mean by standard. – 2017-02-03
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    In the second line, I suppose you mean $\bigoplus_{i \in I}M_i \simeq \bigoplus_{j \in J} N_j$. – 2017-02-03
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    Thats true, but I do still mean product. Since if $(P_i)$ and $(G_i)$ are connected partial orders such that $\amalg_iP_i \simeq \amalg_j G_j$, then there exists a bijection $\sigma: I\rightarrow J$ such that $P_i \simeq G_{\sigma(i)}$. – 2017-02-03
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    Same works for topological spaces: if $(X_i)$ and $(Y_j)$ are connected topological spaces, such that $\amalg_i X_i \simeq \amalg_j Y_j$, then there exists a bijection $\sigma: I\rightarrow J$ such that $X_i \simeq Y_{\sigma(i)}$. – 2017-02-03
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    @StefanPerko - yes. Doing so is often quite useful. However, now that my terminological quibble has been adequately answered. Please let us move on. I have no argument with the question, now that I know what it means. – 2017-02-03
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    I suppose you mean 'indecomposable module', not 'simple module.' – 2017-02-05
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    No I mean simple – 2017-02-05

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The question concerns identifying a class of partial orders whose products have unique factorizations. I will give positive and negative partial answers, both (co)authored by Hashimoto.

The positive answer may be found in

Hashimoto, Junji On direct product decomposition of partially ordered sets. Ann. of Math. (2) 54, (1951). 315-318.

Here it is proved that any two direct product decompositions of a connected poset have a common refinement. Arbitrary products of connected orders are not always connected, but they are sometimes connected. Thus, one can almost take the word simple in the original question to mean connected and directly indecomposable. For example, you can take simple to mean connected and directly indecomposable if you are only concerned about finite products of posets.

The negative result (from an earlier paper of Hashimoto and Nakayama) concerns the nonconnected case. Hashimoto and Nakayama point out that $(1+x)(1+x^2+x^4)=(1+x^3)(1+x+x^2)$, and translate this fact into a statement about posets by letting $1={\bf 1}$ stand for the $1$-element chain, $x={\bf 2}$ stand for the $2$-element chain, multiplication stand for cartesian product, and sum stand for disjoint (parallel) sum.

In other words, there is an isomorphism $A\times B\cong C\times D$ where $A = {\bf 1}+{\bf 2}$ is the disjoint sum of a $1$-element chain and a $2$-element chain, $B={\bf 1}+{\bf 2}^2+{\bf 2}^4$, $C = {\bf 1}+{\bf 2}^3$, and $D = {\bf 1}+{\bf 2}+{\bf 2}^2$. Hashimoto and Nakayama argue that these factorizations have no common refinement.

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    Thank you for your answer, although I think you mean directly irreducible and connected, no? Since connectedness is not enough to get that the factors are pair-wise isomorphic. – 2017-02-05
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    For the non-connected case, just not that any set $X$ together with $=$ is a partial order, and function between sets preserves equality. So if $A = 2$ and $B = 6$ and $C = 4$ and $D = 3$, then $A \times B \simeq C \times D$. – 2017-02-05
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    Re first comment: yes, I will edit. Re second comment: in the Hashimoto-Nakayama example, the factors are directly indecomposable. Not so when the factors are discrete. – 2017-02-05
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    Ok thanks! By the way, do you have a counter-example at hand for why it is the case that an infinite product of connected orders in not necessarily connected? – 2017-02-05
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    Counterexample: Let $P_i$, $i=1,2,3,\ldots$ be a connected poset with distinguished elements $a_i, b_i$ whose shortest connecting path has length $i$. In $\prod P_i$ the elements $(a_i)$ and $(b_i)$ have no connecting path. – 2017-02-05
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    Thanks again for the great answer @Keith Kearnes, it has been very helpful! – 2017-02-10