Solution:

The solution is $74259$. The numbers $74259$, $74592$, and $74925$ are all divisible by $111$. Denote the original number by $\overline{abcde}$ (the line prevents confusion with $a \cdot b \cdot c \cdot d \cdot e$). Then we have
$$\begin{aligned} & 111 \mid \overline{abcde} \\ & 111 \mid \overline{abdec} \end{aligned}$$
Subtracting,
$$\begin{gathered} 111 \mid \overline{abdec} - \overline{abcde} \\ 111 \mid \overline{dec} - \overline{cde} \end{gathered}$$
Since the number $\overline{dec} - \overline{cde} = 90d + 9e - 99c$ is divisible by $9$ we get $33 \mid \overline{dec} - \overline{cde}$. Since it is given that $\overline{abdec} > \overline{abcde}$, $\overline{cde} \leq \overline{dec} - 333$.
However, we could have done the above logic with $d, e, c$ instead of $c, d, e$ and gotten $\overline{dec} \leq \overline{ecd} - 333$. Consequently $\overline{cde} \leq \overline{ecd} - 666$. Since obviously $\overline{ecd} \leq 999$, $\overline{cde}$ is one of the multiples of $37$ up to $333$:
$$000\quad 037\quad 074\quad 111\quad 148\quad 185\quad 222\quad 259\quad 296\quad 333$$
We can immediately eliminate $000, 111, 222$, and $333$, since we know that the digits are all different. We can also eliminate $074, 185$, and $296$, since $\overline{ecd} > \overline{dec}$. The three remaining choices are all of the form $111 \cdot k + 37$. So
$$\begin{aligned} 111 \mid \overline{abcde} & \\ & = \overline{ab000} + \overline{cde} \\ & = 999 \cdot \overline{ab} + \overline{ab} + 111 \cdot k + 37 \\ 111 \mid \overline{ab} + 37 & \end{aligned}$$
whence $\overline{ab} = 74$. This leaves only three possibilities for Wally's combination lock number: $74037$, $74148$, and $74259$, of which only the last has all unlike digits.