(I) Proof: When $n≥2$, $S\_n= \frac {(a\_n+2)(a\_n-1)}{2}(n∈N^{})...①$
$S_{n-1}= \frac {(a_{n-1}+2)(a_{n-1}-1)}{2}...②$

Subtracting ② from ①, we get: $a\_n= \frac {a\_n^{2}+a\_n-a_{n-1}^{2}-a_{n-1}}{2}...(1)$
Simplifying, we get: $(a\_n+a_{n-1})(a\_n-a_{n-1})=(a\_n+a_{n-1})...(2)$

Since all terms of the sequence $\{a\_n\}$ are positive, $a\_n+a_{n-1}≠ 0$,
Hence, $a\_n-a_{n-1}=1(n≥2)...(3)$

When $n=1$, $a\_1=S\_1= \frac {(a\_1+2)(a\_1-1)}{2}$, we get $a\_1^{2}-a\_1-2=0$,
Since $a\_1 > 0$, we get $a\_1=2...(4)$

Hence, the sequence $\{a\_n\}$ is an arithmetic sequence with first term $2$ and common difference $1...(5)$

(II) From $(1)$, we get $a\_n=2+(n-1)×1=n+1...(6)$
Hence, $b\_n=a\_n⋅ 3^{n}=(n+1)⋅ 3^{n}...(7)$

$T\_n=2×3^{1}+3×3^{2}+4×3^{3}+…+n×3^{n-1}+(n+1)×3^{n}...(1)...(8)$
$3T\_n=2×3^{2}+3×3^{3}+4×3^{4}+…+n×3^{n}+(n+1)×3^{n+1}...(2)...(9)$

Subtracting $(2)$ from $(1)$, we get: $-2T\_n=6+3^{2}+3^{3}+…+3^{n}-(n+1)×3^{n+1}...(10)$
Hence, $-2T\_n=6+ \frac {3^{2}-3^{n}×3}{1-3}-(n+1)×3^{n+1}= \frac {3^{n+1}-3}{2}-(n+1)×3^{n+1}...(11)$

Hence, $T\_n= \frac {1}{4}(2n+1)3^{n+1}- \frac {3}{4}...(12)$

The final answers are:
(I) The sequence $\{a\_n\}$ is an arithmetic sequence with first term $\boxed{2}$ and common difference $\boxed{1}$.
(II) The sum of the first $n$ terms of the sequence $\{b\_n\}$, $T\_n$, is $\boxed{\frac {1}{4}(2n+1)3^{n+1}- \frac {3}{4}}$.