Let me reformulate the question a bit and then provide an empirical argument in favor of the negative answer.

For an associative algebra $A$ with involution $J$, define a new binary operation on $A$ as $a \mathbin{\square} b = ab + ba^J$. The question is whether this operation defines a "nice" class of algebras.

This seems to be a correct generalization of the initial question, as by analogy with Lie ($[a,b] = ab - ba$) and Jordan ($a \circ b = ab + ba$) algebras constructed from associative algebras in a similar way, "interesting" and "natural" examples of such algebras do not confined to matrix algebras $A$.

What is a "nice" class of algebras? Again, by analogy with Lie and Jordan situations we might require that it satisfies some nontrivial identity (like Jacobi or Jordan identity), i.e., form a variety of algebras. Then the answer is "no": the class of all associative algebras with involution under the operation $\square$ do not satisfy any nontrivial identity. For, suppose such identity exist. Than each associative algebra $A$ with involution satisfies some identity of the form $f(x_1, \dots, x_n, x_1^J, \dots, x_n^J) = 0$ with nontrivial occurrences of involutions. Then by theorem of Amitsur, $A$ satisfies a power $n$ of a standard identity of degree $d$, both $n$ and $d$ are determined by $f$ (see, for example, I.N. Herstein, Rings with Involution, The Univ. of Chicago Press, 1976, $\S$ 5). As, obviously, not all algebras with involution would satisfy the latter identity, we get a contradiction.

Even if we confine ourselves with matrix algebras with transposition, as in the initial question, we can choose a matrix algebra of sufficiently large degree not satisfying a given power of a standard identity of a given degree, so even that narrower class of algebras do not form a variety.

In fact, this argument is valid for any binary operation defined in terms of addition, multiplication, and involution in an associative algebra, provided involution occurs non-trivially, not necessary $\square$.