Is L closed under linear-time reductions?
L is closed under linear-time reductions if and only if ​L = P.

Proof:

If L is closed under linear-time many-one reductions then:

Since L has non-trivial problems, all problems that are solvable linear time are in L.

Let (x) be the modification of the most basic P-complete problem to ask about "the first sqrt(T) steps" rather than "the first T steps". ​Since unary-to-binary conversion can be done in linear time and [square-roots can be computed and TMs can be simulated in quasi-linear time], (x) can be solved in linear time. ​Since L has non-trivial problems, there is a linear-time many-one reduction from (x) to a problem in L. ​By this paragraph's assumption, that means (x) is in L.  Since L can square unary inputs, there is a logspace reduction from the most basic P-complete problem to (x). ​Thus (x) is P-complete. ​Since (x) is in L, that means ​ P ⊆ L .

If L is closed under linear-time many-one reductions then then ​P ⊆ L. With no assumptions, L ⊆ P. ​ Therefore, if L is closed under linear-time many-one reductions then then​ L = P. Since P can perform linear-time Turing reductions, P is closed under linear-time Turing reductions. Therefore, if ​L = P  then L is closed under linear-time Turing reductions. ​ QED