Perhaps a legend of the symbols used in the two formalizations can be helpful. Can you provide it?

Oh, of course.
Ax := "x is american"
Ex := "x speak english"
Rx := "x is robber"

what about (∀x), ∃x, ⊃, ∧,
I know the latter two to be common math symbols, but I take it most here didn't study math to the extend you seem to have, and even though I tutor math on occasion I will need to look them up to avoid making mistakes.

(∀x) means "to all x", (∃x) means "there exists an x", "⊃" means "implies" and "∧" means "and". So, for instance, the first conclusion (∃x)[Rx∧Ex] would be read as "there exists an x such that x is a robber and x speak english", while the second conclusion (∃x)[Rx⊃Ex] would be read as "there exists an x such that if x is a robber, then x speaks english" (note that this "if ... then" doesn't necessarily means that there is a causation relation between the antecedent and the consequent. It means only that it's not the case that the antecedent is true and the consequent is false)

So for the full list:
A = American
x= people
E= speaks English
R= Robber
∀=all
∃= exists
⊃=implies (that)
∧= and (not and/or)
∴= thus/therefor
Resulting in two expressions:
I)
1. (∀x)[Ax⊃Ex]
2. (∃x)[Rx∧Ax]
3. ∴ (∃x)[Rx∧Ex]
=
1.for all people goes that if they are American that implies they speak English
2. there exists people who are Robbers and are American
3. thus there exist Robbers that speak English

II)
1. (∀x)[Ax⊃Ex]
2. (∃x)[Rx⊃Ax]
3. ∴ (∃x)[Rx⊃Ex]
=
1.for all people goes that if they are American and that implies they speak English
2. there exist people who are Robbers and that implies they are American
3. thus there exist people who are Robbers and that implies they speak English

Assuming I understood correctly the answer is rather obvious. presupposition 2 in case II is different from presupposition 2 in case I. Case I presupposition 2 seems reasonable, but case II presupposition 2 does not since it excludes the existence of Robbers not being American which obviously isn't the case. Hence the conclusion in case II is invalid while the conclusion in case I stands.
So the answer to your question is: No, the formulation II is invalid, assuming I understood correctly.

Though it's rather confusing since the math symbol ⊃ has two uses. If restricting yourself to the meaning of '⊃ = implies' my conclusion stands. However since ⊃ in math can also mean subset (as in A⊃B means every element of B is also element of A while A ≠ B (since A consists of more elements than B), then a different conclusion might be the case.

Hence my personal preference to use => or --> rather than ⊃ to mean imply, and usually I only use ⊃ to mean subset when it comes to math notations.

In short the statement:
" 2. Some robbers are americans"
is not correctly represented by the statement
"2. (∃x)[Rx⊃Ax]"
since the latter statement make a claim about all robbers and not just some robbers.

But the second premise in the argument II does not excludes the existence of robbers not being american. It says only that exists at least one robber that is american. It doesn't says nothing about robbers that aren't americans.

Robber implies American, means if robber then American, thus all robbers are American though not nessesarily all americans are robbers.
quoted from: https://en.wikipedia.org/wiki/List_of_mathematical_symbols
A ⇒ B means if A is true then B is also true; if A is false then nothing is said about B.
(→ may mean the same as ⇒, or it may have the meaning for functions given below.)
(⊃ may mean the same as ⇒,[12] or it may have the meaning for superset given below.)

hence if you state R => A you are excluding the possibility of a Robber to be not American. Perhaps you didn't intend it to mean this, but that is the mathematical convention. As you noted Rx ⊃ Ax, that means that if for person x R is true, then for that person x A is true as well. Excluding the possibility that if for person x R is true A is not true.

Not to be confused with the subset usage of the symbol ⊃ :
A ⊇ B means every element of B is also an element of A.
A ⊃ B means A ⊇ B but A ≠ B.

Well... No. As you quoted, if A is false, nothing is said about B.
But you need to notice that it is a existential claim. You are treating it like as if it were a propositional logic claim, like "if it's a robber, then it's american" (if it were the case, then you would be right), but it's an existential predicative logic claim. It is said that there exists an x such that If it's a robber, then it's american. (not simply that "robber implies american"). It doesn't means that all robbers are americans, but that there are at least one. It can be the case that all robbers are americans, but it can be the case that there are robbers that aren't americans. If there is an american robber and a mexican one, the claim "(∃x)[Rx⊃Ax]" is also true. You are commiting a logical jump by infering an universal claim from an existential one.

too late here, I'm brabbling nonsense. Need sleep first.