Are one-by-one matrices equivalent to scalars?
No. To give a concrete example, you can multiply a 2x2 matrix by a scalar, but you can't multiply a 2x2 matrix by a 1x1 matrix.
It is sloppy notation.
Some Background
There are three basic kinds of mathematical spaces that are being asked about in the original question: scalars, vectors, and matrices. In order to answer the question, it is perhaps useful to develop a better understanding of what each of these objects is, or how they should be interpreted.
In this context, a scalar is a an element of a field. Without getting into too much detail, a field is an object made from three ingredients $(k,+,\cdot)$, where $k$ is a set, and $+$ and $\cdot$ are two binary operations (addition and multiplication). These operations are associative and commutative, there are distinguished identity elements (0 and 1) for each operation, every element has an additive inverse, every nonzero element has a multiplicative inverse, and multiplication distributes over addition.
Examples of fields include (among others) the real numbers, as well as the rationals, the complex numbers, and (if you are feeling a little more esoteric) the $p$-adic numbers.
A vector is an element of a vector space. A vector space is an object made from four ingredients $(V,k,+,\cdot)$, where $V$ is a set, $k$ is a field (the base field), $+$ is a binary operation which acts on two vectors, and $\cdot$ is a binary operation (called scalar multiplication) which acts on a scalar from $k$ and a vector from $V$. The addition is associative and commutative, there is a distinguished 0 vector (the additive identity), and every vector has an additive inverse. Scalar multiplication allows scalars to "act on" vectors; a vector can be multiplied by a scalar, and this multiplication "plays nice" with the addition (e.g. $a(b\mathbf{v}) = (ab)\mathbf{v}$ and $a(\mathbf{u}+\mathbf{v}) = a\mathbf{u} + a\mathbf{v}$ for scalars $a$ and $b$, and vectors $\mathbf{u}$ and $\mathbf{v}$).
Examples of vector spaces include $\mathbb{R}^n$ (the base field is $k = \mathbb{R}$), $\mathbb{C}^n$ (the base field is $\mathbb{C}$), and the space of continuous functions from $[0,1]$ to $\mathbb{C}$ (i.e. the space $C([0,1])$–the base field here is $\mathbb{C}$ again).
Finally, a matrix is an element of a specific kind of algebra over a field. An algebra has two ingredients: $(V,\times)$, were $V$ is a vector space and $\times$ is a binary operation called a bilinear product. This product is both left- and right-distributive over the addition in $V$, and plays nice with scalar multiplication. In detail, if $a$ and $b$ are scalars and $\mathbf{u}$, $\mathbf{v}$, and $\mathbf{w}$ are vectors, then
- $\mathbf{u} \times (\mathbf{v} + \mathbf{w}) = (\mathbf{u}\times \mathbf{v}) + (\mathbf{u}\times \mathbf{w})$,
- $(\mathbf{u} \times \mathbf{v}) \times \mathbf{w} = (\mathbf{u}\times \mathbf{w}) + (\mathbf{v}\times \mathbf{w})$, and
- $(a\mathbf{u})\times (b\mathbf{v}) = (ab)(\mathbf{u}\times \mathbf{v}).$
Note that the underlying vector space (with its underlying field) is still running around, so there is a ton of structure here to play around with.
Examples of algebras include $\mathbb{R}^3$ with the usual cross product, and matrix algebras such as the space of $n\times n$ matrices over $\mathbb{R}$ with the usual matrix multiplication. The structure of an algebra is actually more general than this, and there are much more interesting examples (such as $C^{\ast}$-algebras), but these are not really germane to this question.
The Question Itself
Now to attempt to answer the question:
If $k$ is any field, then we can regard $k$ as a (one-dimensional) vector space over itself by taking the scalar multiplication to coincide with the multiplication in the field. That is, if we let $a$ denote a scalar and $\langle a \rangle$ denote a vector, then $$ a \cdot \langle b \rangle := \langle a\cdot b \rangle. $$ In a mathematically meaningful sense, this is a way of identifying the field $(k,+,\cdot)$ with the vector space $(V,k,+,\cdot)$ (though something is lost in the translation–it doesn't make sense to multiply two vectors, though it does make sense to multiply a vector by a scalar). This is a bit outside my area of expertise, but I think that the right language (from category theory) is that there is an faithful functor from the category of fields to the category of vector spaces. In more plain language, this says that fields and (certain) vector spaces are equivalent to each other. Specifically, we can sometimes regard a one-dimensional vector as a scalar–while they are not quite the same objects, they can, in the right context, be treated as though they are the same. In this sense, a scalar is "functionally equivalent" to a one-dimensional vector.
By a similar argument, we can regard a field $k$ as an algebra over itself, with the underlying vector space being that obtained by viewing $k$ as a vector space, and the bilinear product being the product obtained by making the identification $$ [a]\times [b] = [a\cdot b]. $$ In this case, the algebra obtained is the algebra of $1\times 1$ matrices over $k$. Hence there is a faithful function identifying any field with an algebra of $1\times 1$ matrices. Again, this gives us a way of identifying scalars with matrices, so (again) we may meaningfully assert that scalars can be identified with matrices. In contrast to the previous example, we really don't lose anything in the translation. Even more surprising, the bilinear product ends up being a commutative operation, which is a neat property for an algebra to have.
It is worth observing that, in this setting, $[a]$ is not "the scalar $a$ in some kind of container." The notation $[a]$ denotes a $1\times 1$ matrix with entry $a$. The brackets denote a great deal of structure–more than is implied by the simple statement "a scalar in a container."
Long Answer Short: A $1\times 1$ matrix is not a scalar–it is an element of a matrix algebra. However, there is sometimes a meaningful way of treating a $1\times 1$ matrix as though it were a scalar, hence in many contexts it is useful to treat such matrices as being "functionally equivalent" to scalars. It might be a little sloppy to do so, but a little bit of sloppiness is forgivable if it introduces no confusion or ambiguity, and if it aids brevity or clarity of exposition.