How to replace all elements of a list by a rule "element" -> "element_"

Here is one way:

Pattern[#, Blank[]] & /@ {a, b, c, d, e, f, g, h, i, j}
(* {a_, b_, c_, d_, e_, f_, g_, h_, i_, j_} *)

An inspection of the FullForm of a_ reveals why this works:

a_ // FullForm
(* Pattern[a, Blank[]] *)

We can abbreviate slightly if we realize that the InputForm of Blank[] is _:

Pattern[#, _] & /@ {a, b, c, d, e, f, g, h, i, j}
(* {a_, b_, c_, d_, e_, f_, g_, h_, i_, j_} *)

As an alternative approach, one might think to use pattern-matching replacement instead:

Replace[{a, b, c, d, e, f, g, h, i, j}, s_Symbol :> s_, {1}]
(*
  RuleDelayed::rhs: Pattern s_ appears on the right-hand side of rule s_Symbol:>s_.
  {a_, b_, c_, d_, e_, f_, g_, h_, i_, j_}
*)

... but Mathematica issues a warning because most of the time having a pattern on the right-hand side of a rule is a mistake. In this particular case it is not an error, so we have to use Quiet to tell Mathematica that:

Quiet[Replace[{a, b, c, d, e, f, g, h, i, j}, s_Symbol :> s_, {1}], RuleDelayed::rhs]
(* {a_, b_, c_, d_, e_, f_, g_, h_, i_, j_} *)

A few additional alternatives to inject patterns on the rhs:

list = {a, b, c, d, e, f, g, h, i, j};

Replace[list, a_ :> (x_ /. x -> a), 1]

{a_, b_, c_, d_, e_, f_, g_, h_, i_, j_}

Replace[list, a_ :> (Pattern[#, _] &@a), 1]

{a_, b_, c_, d_, e_, f_, g_, h_, i_, j_}

Activate @ Replace[list, a_ :> Inactive[Pattern][a, _], 1]

{a_, b_, c_, d_, e_, f_, g_, h_, i_, j_}

Replace[list, a_ :> foo[a, _], 1] /. foo -> Pattern

{a_, b_, c_, d_, e_, f_, g_, h_, i_, j_}

convert it to string and then deal with it

ToExpression@StringJoin[#,"_"]&/@ToString/@{a,b,c,d,e}

a shorter way

ToExpression@StringJoin[ToString@#,"_"]&/@{a,b,c,d,e}