But if you do have independence, the proof is easy! Let S be the sum of everyone else and X be the discrete uniform random in [0, 9]. Then:
Pr(X + S mod 10 = i)
= \sum_j Pr((X + j) mod 10 = i | S = j) Pr(S = j)
= \sum_j Pr(X = i | S = j) Pr(S = j)
= \sum_j Pr(X = i) Pr(S = j)
= Pr(X = i) \sum_j Pr(S = j)
= Pr(X = i)
Due to total probability, symmetry, independence, and total probability again respectively. The handwaved part is the mod where you can imagine the histogram columns in the pmf getting rotated/shifted around but ending up looking exactly the same afterwards since all columns are symmetrical.