I'm afraid that your definition of a junk theorem is somewhat imprecise since it doesn't address what the language of the theory $T$ is and how it is interpreted in the language of set theory. For one thing there are surely lots of theorems we would like to categorize as "junk" which are not even expressible in the language of $T$. Another issue is that a sentence like $5 \cap 17 = 5$ need not be junk depending on how you interpret your theory in set theory. For example if $T$ is the theory of lattices, then you can interpret $\omega$ (the set of finite von Neumann ordinals)...

Here's a random thought that occurred to me: wouldn't it make sense to consider spaces which are disjoint unions of simply-connected spaces? The content of homotopy theory should be essentially the same, but maybe the category of such spaces would be better behaved (for one thing it has coproducts)?

I took a closer look at the proof I mentioned before. While it seems that it indeed avoids transversality, it still uses smoothness in an essential way. So even though my original question is now answered even in the topological case, the excision axiom still poses a problem. And of course while we can leave without the weak equivalence axiom, the excision is really essential...

This answer makes everything clear, thank you. If I'm not mistaken the claim you prove in your second to last paragraph could be restated by saying that $\partial M \to M$ is a mixed cofibration in the sense of Cole. I wonder if your last paragraph has some nice interpretation in these terms.

However, is it true that $\partial M \to M$ is a Serre cofibration when $M$ is merely a topological manifold? For the above argument knowing that it is a Hurewicz cofibration would not be sufficient.

OK, I think I figured out what you meant in the smooth case. The inclusion $\partial M \to M$ is a Serre cofibration for a smooth $M$. Thus if we have a weak equivalence $p : X \to Y$, singular manifolds $f : M \to X$, $g : N \to X$ and a bordism $H : W \to Y$ from $(M, f)$ to $(N, g)$, then we can find a lift $K : W \to X$ so that $K$ is bordism from $(M, p f)$ to $(N, p g)$ (the other triangle will only commute up to homotopy). This shows injectivity of $MO_* X \to MO_* Y$ and surjectivity is even easier.

@Ricardo: I find it plausible that those observations could be useful in the proof, but could you perhaps sketch how they are used in the argument? I don't see it myself.

so I don't see how the axiom follows. Lennart Meier suggested to me that it could be verified by looking at a handle decomposition of $M$, but I don't see the details yet (and it would still pose issues for the topological case).

@Tom: there is no issue with excision. In the book I mentioned Conner and Floyd prove it without transversality, they use "straightening of angles" which roughly amounts to the observation that if you take the product of two smooth manifolds with boundary, then you can "smooth the corners" so that it becomes a smooth manifold with boundary itself. As to your other remark, it is indeed true that a weak equivalence $X \to Y$ induces a bijection $[M, X] \to [M, Y]$ for any manifold $M$. However the bordism relation identifies more than the homotopy relation...

It's hard to say whether it addresses your question, but you may try to take a look at Mather Hurewicz Theorems for Pairs and Squares.

The case $I = J = [1]$ is indeed easy. If $X \to Y$ is a morphism in $\mathcal{C}^{[1]}$ and $Z$ is the corresponding object of $\mathcal{C}^{[1] \times [1]}$, then we have $L_{1,1} Z = Z_{1,0} \sqcup_{Z_{0,0}} Z_{0,1} = X_1 \sqcup_{X_0} Y_0 = X_1 \sqcup_{L_1 X} L_1 Y$. Thus the canonical morphism $L_{1,1} Z \to Z_{1,1}$ coincides with $X_1 \sqcup_{L_1 X} L_1 Y \to Y_1$ and this morphism being a cofibration is equivalent to both $Z$ being Reedy cofibrant and $X \to Y$ being a Reedy cofibration (assuming that we already verified this at $(0,0)$, $(1,0)$ and $(0,1)$, which is easier).

Immediately after posting my answer I regretted using the word "easily" twice... Addressing your questions will require editing the answer, it will take a while.