Find two algebras $A,B$ such that neither can be embedded into $A \times B$
This appears to be really subtle. My initial thoughts are letting $A = B =$ the one element Boolean algebra where $0 = 1$. It seems like that $A \times B$ would give something that pictorially looks like the the 2-element Boolean algebra but that neither $A$ or $B$ would embed into this because of the property $1 = 0$.
I can't really think of any groups or other types of algebras where this may fail.
Given an embedding $\iota\colon A\to A\times B$, you can compose it with the projection homomorphism $\pi\colon A\times B\to B$ to get a homomorphism $\varphi = \pi\circ\iota \colon A\to B$. Conversely, if there exists a homomorphism $\varphi\colon A\to B$, then there is an induced homomorphism $(\textrm{id}_A\times \varphi)\colon A\to A\times B$ given by the universal property of products, which is an embedding since the first coordinate map is. This shows that there exists an embedding of $A$ into $A\times B$ iff there is a homomorphism from $A$ to $B$.
This makes it easy to construct examples where both, neither, or exactly one of $A$ and $B$ are embeddable in $A\times B$. E.g., consider the products of commutative unital rings $\mathbb Z_2\times \mathbb Z_2$ (both factors embeddable), $\mathbb Z_2\times \mathbb Z_3$ (neither factor embeddable), and $\mathbb Z_2\times \mathbb Z_4$ (only second factor is embeddable).