Briefly: there's a simple difference in how they treat 0. That fixed, still neither implies the other in general. In a regular extensive category, a slight modification of the LS definition implies the Elephant one. I suspect they're not fully equivalent in anything short of a topos.

The simple difference: 0 is always indecomposable by Lambek and Scott's definition (since any map into 0 is epi), but never by the Elephant's (since the uniqueness condition won't hold; or by considering when the coproduct decomposition is empty). So, let's temporarily change one of the definitions to fix this. I'd suggest we add “…and the map $0 \to X$ is not epi.” to Lambek and Scott's definition. (As you noted, their binary condition generalises to a $k$-ary one; this is just the case $k=0$.)

In eg Top, however, we can see that the Elephant def still doesn't imply the LS def. $[0,1]$ satisfies the former (it's not decomposable by an iso), but not the latter (it is decomposable by an epi). Even more, it’s decomposable by a regular epi (more on this distinction below).

Conversely, the LS definition doesn't imply the Elephant one either; it fails in eg $\mathbf{Set}^\mathrm{op}$, since in $\mathbf{Set}$, $0$ is co-decomposable by iso ($0 \cong A \times 0$) but not co-decomposable by monos (for any map $(f,g) \colon 0 \to A \times B$, not just one but both of $f$ and $g$ are mono).

When do they imply each other? If we upgrade the LS definition to involve regular epis, then in a regular lextensive category, it implies the Elephant definition, if I'm not mistaken. For this, suppose $X$ is “indecomposable by reg epis”, and suppose $X \cong A + B$ — WLOG $X = A + B$. The coproduct inclusions are then jointly reg epi, so one of them is reg epi. But it's also mono (in a lextensive category, every coproduct inclusion is a pullback of $1 \to 1 + 1$, so is mono); so it's iso. There's a little more fiddly stuff to check involving messing around with $0$, but it's all the same sort of thing.

If we again look at just regular epis, I'd guess they're completely equivalent in a pretopos (I'd imagine this might be nice to show by working in the internal logic). Without adding “regular”, I wouldn't imagine they're fully equivalent until we get to a topos, where all epis are regular — just as an instance of the general principle that not-necessarily-regular epis very rarely well-behaved.

Briefly: there's a simple difference in how they treat 0. That fixed, still neither implies the other in general. In a regular extensive category, a slight modification of the LS definition implies the Elephant one. I suspect they're not fully equivalent in anything short of a topos. As Mike Shulman points out, even in a topos they are not equivalent.

The simple difference: 0 is always indecomposable by Lambek and Scott's definition (since any map into 0 is epi), but never by the Elephant's (since the uniqueness condition won't hold; or by considering when the coproduct decomposition is empty). So, let's temporarily change one of the definitions to fix this. I'd suggest we add “…and the map $0 \to X$ is not epi.” to Lambek and Scott's definition. (As you noted, their binary condition generalises to a $k$-ary one; this is just the case $k=0$.)

In eg Top, however, we can see that the Elephant def still doesn't imply the LS def. $[0,1]$ satisfies the former (it's not decomposable by an iso), but not the latter (it is decomposable by an epi). Even more, it’s decomposable by a regular epi (more on this distinction below).

Conversely, the LS definition doesn't imply the Elephant one either; it fails in eg $\mathbf{Set}^\mathrm{op}$, since in $\mathbf{Set}$, $0$ is co-decomposable by iso ($0 \cong A \times 0$) but not co-decomposable by monos (for any map $(f,g) \colon 0 \to A \times B$, not just one but both of $f$ and $g$ are mono).

When do they imply each other? If we upgrade the LS definition to involve regular epis, then in a regular lextensive category, it implies the Elephant definition, if I'm not mistaken. For this, suppose $X$ is “indecomposable by reg epis”, and suppose $X \cong A + B$ — WLOG $X = A + B$. The coproduct inclusions are then jointly reg epi, so one of them is reg epi. But it's also mono (in a lextensive category, every coproduct inclusion is a pullback of $1 \to 1 + 1$, so is mono); so it's iso. There's a little more fiddly stuff to check involving messing around with $0$, but it's all the same sort of thing.

Edit from Mike Shulman's comments: if moreover we're in a pretopos, all epis are regular, so there the original LS definition will imply the Elephant definition. On the other hand, the Elephant definition doesn't imply the LS even in a topos: the terminal object of $\mathbf{Sh}([0,1])$ is a counterexample, essentially for the same reasons that $[0,1]$ was a counterexample in $\mathbf{Top}$.

However, the two definitions are equivalent for projective objects… and I guess that's how this situation has arisen, since a common use of indecomposable objects in topos theory is the theorem that the indecomposable projectives in a presheaf category are exactly the retracts of representables. (This is useful because it lets us recover the idempotent-completion of $\mathbf{C}$, which is very close to $\mathbf{C}$ itself, from $[\mathbf{C}^\mathrm{op},\mathbf{Set}]$.)

Briefly: there's a simple difference in how they treat 0. That fixed, still neither implies the other in general. In a regular extensive category, a slight modification of the LS definition implies the Elephant one. I suspect they're not fully equivalent in anything short of a topos.

The simple difference: 0 is always indecomposable by Lambek and Scott's definition (since any map into 0 is epi), but never by the Elephant's (since the uniqueness condition won't hold; or by considering when the coproduct decomposition is empty). So, let's temporarily change one of the definitions to fix this. I'd suggest we add “…and the map $0 \to X$ is not epi.” to Lambek and Scott's definition. (As you noted, their binary condition generalises to a $k$-ary one; this is just the case $k=0$.)

In eg Top, however, we can see that the Elephant def still doesn't imply the LS def. $[0,1]$ satisfies the former (it's not decomposable by an iso), but not the latter (it is decomposable by an epi). Even more, it’s decomposable by a regular epi (more on this in a moment).

Conversely, the LS definition doesn't imply the Elephant one either; it fails in eg $\mathbf{Set}^\mathrm{op}$, since in $\mathbf{Set}$, $0$ is co-decomposable by iso ($0 \cong A \times 0$) but not co-decomposable by monos (for any map $(f,g) \colon 0 \to A \times B$, not just one but both of $f$ and $g$ are mono).

When do they imply each other? If we upgrade the LS definition to involve regular epis, then in a regular lextensive category, it implies the Elephant definition, if I'm not mistaken.  (Have to go now; can give details of this later.)

If we again look at just regular epis, I'd guess they're completely equivalent in a pretopos (I'd imagine this should be nice to show by working in the internal logic). Without adding “regular”, I don't imagine they're fully equivalent until we get to a topos, where all epis are regular.

Briefly: there's a simple difference in how they treat 0. That fixed, still neither implies the other in general. In a regular extensive category, a slight modification of the LS definition implies the Elephant one. I suspect they're not fully equivalent in anything short of a topos.

The simple difference: 0 is always indecomposable by Lambek and Scott's definition (since any map into 0 is epi), but never by the Elephant's (since the uniqueness condition won't hold; or by considering when the coproduct decomposition is empty). So, let's temporarily change one of the definitions to fix this. I'd suggest we add “…and the map $0 \to X$ is not epi.” to Lambek and Scott's definition. (As you noted, their binary condition generalises to a $k$-ary one; this is just the case $k=0$.)

In eg Top, however, we can see that the Elephant def still doesn't imply the LS def. $[0,1]$ satisfies the former (it's not decomposable by an iso), but not the latter (it is decomposable by an epi). Even more, it’s decomposable by a regular epi (more on this distinction below).

Conversely, the LS definition doesn't imply the Elephant one either; it fails in eg $\mathbf{Set}^\mathrm{op}$, since in $\mathbf{Set}$, $0$ is co-decomposable by iso ($0 \cong A \times 0$) but not co-decomposable by monos (for any map $(f,g) \colon 0 \to A \times B$, not just one but both of $f$ and $g$ are mono).

When do they imply each other? If we upgrade the LS definition to involve regular epis, then in a regular lextensive category, it implies the Elephant definition, if I'm not mistaken. For this, suppose $X$ is “indecomposable by reg epis”, and suppose $X \cong A + B$ — WLOG $X = A + B$. The coproduct inclusions are then jointly reg epi, so one of them is reg epi. But it's also mono (in a lextensive category, every coproduct inclusion is a pullback of $1 \to 1 + 1$, so is mono); so it's iso. There's a little more fiddly stuff to check involving messing around with $0$, but it's all the same sort of thing.

If we again look at just regular epis, I'd guess they're completely equivalent in a pretopos (I'd imagine this might be nice to show by working in the internal logic). Without adding “regular”, I wouldn't imagine they're fully equivalent until we get to a topos, where all epis are regular — just as an instance of the general principle that not-necessarily-regular epis very rarely well-behaved.