$_[0] is not a copy. It's literally the same scalar that was passed as an argument. That means the first 1+1 produced a read-only scalar and the second one didn't.

We can see some differences in the generated code.

$ perl -MO=Concise,-exec -MDevel::Peek -we'Dump 1; sub foo{Dump $_[0];
+$_[0]=1} foo(1)'
1  <0> enter
2  <;> nextstate(main 44 -e:1) v:{
3  <$> const[IV 1] s
4  <2> Devel_Peek_Dump vK/1
5  <;> nextstate(main 46 -e:1) v:{
6  <0> pushmark s
7  <$> const[IV 1] sM
8  <#> gv[IV \&main::foo] s
9  <1> entersub vKS
a  <@> leave[1 ref] vKP/REFC
-e syntax OK
[download]

$ perl -MO=Concise,-exec -MDevel::Peek -we'Dump 1+1; sub foo{Dump $_[0
+];$_[0]=1} foo(1+1)'
1  <0> enter
2  <;> nextstate(main 44 -e:1) v:{
3  <$> const[IV 2] s/FOLD
4  <2> Devel_Peek_Dump vK/1
5  <;> nextstate(main 46 -e:1) v:{
6  <0> pushmark s
7  <$> const[IV 2] sM/FOLD
8  <#> gv[IV \&main::foo] s
9  <1> entersub vKS
a  <@> leave[1 ref] vKP/REFC
-e syntax OK
[download]

So it appears that a folded constant (FOLD) in lvalue context (M) returns a writable scalar.

We can also see the difference when using the referencing operator or foreach, as they also also impose an lvalue context.

$ perl -MDevel::Peek -e'Dump(${\1})'
SV = IV(0x8549e4010) at 0x8549e4020
  REFCNT = 1
  FLAGS = (IOK,READONLY,PROTECT,pIOK)
  IV = 1

$ perl -MDevel::Peek -e'Dump(${\(1+1)})'
SV = IV(0x7ed5d4d610) at 0x7ed5d4d620
  REFCNT = 1
  FLAGS = (IOK,pIOK)
  IV = 2

$ perl -MDevel::Peek -e'Dump($_) for 1'
SV = IV(0x151c4b8240) at 0x151c4b8250
  REFCNT = 2
  FLAGS = (IOK,READONLY,PROTECT,pIOK)
  IV = 1

$ perl -MDevel::Peek -e'Dump($_) for 1+1'
SV = IV(0x567114cd90) at 0x567114cda0
  REFCNT = 1
  FLAGS = (IOK,pIOK)
  IV = 2
[download]

I believe this is intentional. Addition normally produces a writable scalar, and those semantics are being preserved even when the operands of the addition are constants.