Multiplying by a $1\times 1$ matrix? For matrix multiplication to work, you have to multiply an $m \times n$ matrix by an $n \times p$ matrix, so we have $$\bigg(m \times n\bigg)\bigg(   n\times p \bigg).$$

But what about a $1 \times 1$ matrix? Is this just a scalar? But every matrix can be mulitplied by a scalar; so do $1 \times 1$ matrices break the rule?
A $1\times 1$ matrix is not the same thing as a scalar.  There is a one to one correspondence between them, but $[a]$ and $a$ are two different things, the first being a matrix and the second one being a scalar.  The confusion comes from the fact that the rings $R$ and $M_1(R)$ are isomorphic, and so $M_n(R)$ is isomorphic to $M_n(M_1(R))$.  When you talk about treating $[a]$ as a scalar, you are implicitly using this isomorphism.
Yes!  The simple answer is that a 1 by 1 matrix is a scalar and a scalar is a one by one matrix.  This makes sense because if you regard the dot product of two vectors (which always returns a scalar) as a row vector times a column vector then you always get a 1 by 1 matrix.  There is no distinction as they have all the same properties (commutativity, same inverse, associative, etc...), if you want to be "weird", you can leave the [brackets] around the 1 by 1 matrix and say that there is a isomorphism from the 1 by 1 group of matrices to the scalars of F (R or C, etc).   Define it as, g([A])=A.  If you want to be really neat you need to just check that if the matrix is [0] that it has no inverse, and therefore g([0])=0 and g^-1 does not exist for this particular "matrix".