Let $X,Y$ be some nice measurable spaces (I'm interested in $[0,1]$ so we can assume compact, etc.). Let $\mu$ be a measure on $X\times Y$ (again, assume it's a nice probability measure, or even something strong if needed).
Define $\mu_1(A)=\mu(A\times Y)$ and reps. $\mu_2$.
I imagine that Fubini's theorem should hold, meaning that: $\mu(A)=\int\mu_1(A^y)d\mu_2(y)$? is that necessarily true?
Could you give some type of conditions under which it is true?
If I were to prove it, I would probably go to the original proof of Fubini and try to manipulate it, although it would be a long job.
A type of conditions I'm okay with allowing $\mu$ to have: Assume $\mu_1$ is invariant under $x\to(x+a) \operatorname{mod} 1$for all $a$ (or just assume it's Lebesgue!!), and $\mu$ is invariant under $(x,y)\to(x,y+a)\mod 1$ for all $a$ (in particular $\mu_2$ is also Lebesgue, but we have something strong too). Could you deduce that Fubini holds in the manner that I suggested? (which would give the desired $\mu$ is Lebesgue)
(Focusing on the case $X=Y=[0,1]$ with Borel sets.)
1. A probability measure on $[0,1]\times[0,1]$ admits a disintegration $$ \mu(A)=\int_{[0,1]}\mu_1(dx)K(x,A),\qquad A\in\mathcal B([0,1]^2), $$ with $K$ a kernel such that $x\mapsto K(x,A)$ is Borel measurable for each $A\in \mathcal B([0,1]^2$ and $A\mapsto K(x,A)$ is a probability measure on $[0,1]^2$ for each $x\in[0,1]$. Moreover, $K$ can be chosen so that $\mu_1\{x:K(x,x\times[0,1])=1)=1$. The kernel $K$ is as unique as you could reasonably wish for: If $K'$ is another kernel with the properties listed above, then $\mu_1(x:K(x,\cdot)\not=K'(x,\cdot))=0$.
Of course, $\mu$ is the product of its marginals ($\mu=\mu_1\otimes\mu_2$) if and only if $K(x,A)=\mu_2(A_x)$ for $\mu_1$-a.e. $x\in[0,1]$.
(An analogous disintegration with respect to the second marginal is also available.)
2. The special case raised in your comment can be shown to hold using disintegration. A direct argument is also available. Thus, suppose that $\mu_1$ in invariant under translations mod 1, and that $\mu$ is invariant under "vertical" translations mod 1. As you note, we then have $\mu_1=\lambda$ (Lebesgue measure). Let $F:[0,1]\times[0,1]\to\Bbb R$ be Borel measurable and bounded. Then (additions below understood to be modulo 1) $$ \int_{[0,1]^2}F\,d\mu=\int_{[0,1]^2}F(x,y+a)\,\mu(dx,dy). $$ Now integrate both sides of this with respect to $a\in[0,1]$, using Fubini $$ \int_{[0,1]^2}F\,d\mu=\int_{[0,1]^2}G(x)\mu(dx,dy)=\int_{0,1]}G(x)\mu_1(dx)=\int_0^1 G(x)\,dx. $$ here $G(x)=\int_0^1 F(x,y+a)\,da = \int_0^1 F(x,y)\,dy$. It follows that $$ \int_{[0,1]^2}F\,d\mu=\int_0^1 \left[\int_0^1 F(x,y)\,dy\right]\,dx. $$ That is, $\mu$ is $\lambda\otimes\lambda$.