I will prove the following stronger statement by induction on $k$.
Claim: if $R$ is any ring (commutative with $1$), $x_1, \dots, x_k$ is a regular sequence in $R$, and we are given integers $0 \leq d_i \leq e_i$ for all $1 \leq i \leq k$, then we have
$$ (x_1^{e_1}, \dots, x_k^{e_k}) : (x_1^{d_1}, \dots, x_k^{d_k}) = (x_1^{e_1}, \dots, x_k^{e_k}, x_1^{e_1 - d_1} \cdots x_k^{e_k - d_k}). $$
(Of course this recovers your claim by taking $R$ to be of characteristic $p$ with all $e_i = p$ and all $d_i = 1$.)
Step 0: Setup
For brevity, we will write the three ideals above as $I, J$, and $K$ respectively; we want to show $I : J = K$. We clearly have $K \subseteq I : J$. For the reverse containment, we induct on $k$; the base case $k=1$ is trivial. Suppose we have an integer $k > 1$ such that the statement is true for regular sequences of length $k-1$ (in any ring), and fix a regular sequence $x_1, \dots, x_k$ in some ring $R$. Recall that then $x_1^{f_1}, \dots, x_k^{f_k}$ is also a regular sequence for any integers $f_i \geq 1$.
Suppose we are given an element $r$ of $I : J$; that is, $r \in R$ such that $r x_i^{d_i} \in I$ for all $i$. We must show that $r \in K$; as a warning, I will repeatedly modify $r$ by elements of $K$ in the course of the proof.
Step 1: first use of inductive hypothesis
Our assumption implies that the image of $r$ in $R/(x_1^{e_1})$ lies in the colon ideal
$$
(x_2^{e_2}, \dots, x_k^{e_k}) : (x_2^{d_2}, \dots, x_k^{d_k}),
$$
which by the inductive hypothesis equals
$$
(x_2^{e_2}, \dots, x_k^{e_k}, x_2^{e_2 - d_2} \cdots x_k^{e_k - d_k}).
$$
Lifting back up to $R$, we have $r \in (x_1^{e_1}, x_2^{e_2}, \dots, x_k^{e_k}, x_2^{e_2 - d_2} \cdots x_k^{e_k - d_k})$. Since all but the last generator of this ideal lie in $K$, we may adjust $r$ to assume without loss of generality that $r \in (x_2^{e_2 - d_2} \cdots x_k^{e_k - d_k})$. So set $r = ax_2^{e_2 - d_2} \cdots x_k^{e_k - d_k}$. To show that $r \in K$, it will suffice to show that $a \in (x_1^{e_1 - d_1}, x_2^{d_2}, \dots, x_k^{d_k})$.
Step 2: calculations with $r x_1^{d_1}$
The assumption $r \in I : J$ tells us in particular that $r x_1^{d_1} \in I$; that is, that
$$ \tag{1}
a x_1^{d_1} x_2^{e_2 - d_2} \cdots x_k^{e_k - d_k} = b_1 x_1^{e_1} + \cdots + b_k x_k^{e_k}
$$
for some $b_i \in R$. Our next goal is to simplify equation (1) slightly by removing its last term, at the cost of modifying $a$ (and thus $r$) as well as $b_1, \dots, b_{k-1}$.
To accomplish this, we set
\begin{align*}
I' & = (x_1^{e_1}, \dots, x_{k-1}^{e_{k-1}}), \\
I_1' & = I' + (x_1^{d_1}), \\
I_2' & = I' + (x_2^{e_2 - d_2}), \\
& \dots \\
I_{k-1}' & = I' + (x_{k-1}^{e_{k-1} - d_{k-1}}),
\end{align*}
and consider equation (1) separately modulo each of the ideals $I_1', \dots, I_{k-1}'$. Working modulo $I_1'$ gives
$$
0 \equiv 0 + \cdots + 0 + b_k x_k^{e_k} \pmod{I_1'},
$$
which implies that $b_k \in I_1'$. (Reason: $x_k^{e_k}$ is a non-zero-divisor mod $I_1'$, because $x_1^{d_1}, x_2^{e_2}, \dots, x_{k-1}^{e_{k-1}}, x_k^{e_k}$ is a regular sequence--unless one of the exponents is $0$, in which case $I_1'$ is the unit ideal and thus it automatically contains $b_k$.) The analogous calculations modulo $I_2', \dots, I_{k-1}'$ show that $b_k$ lies in each of these ideals.
Now, since $b_k$ lies in each of the ideals $I_1', \dots, I_{k-1}'$, it follows that multiplying it by any of the elements
$$
x_1^{e_1 - d_1}, x_2^{d_2}, \dots, x_{k-1}^{d_{k-1}}
$$
yields an element of $I'$; that is, $b_k$ lies in the colon ideal
$$
I' : (x_1^{e_1 - d_1}, x_2^{d_2}, \dots, x_{k-1}^{d_{k-1}}).
$$
Our inductive hypothesis implies that this ideal equals
$$
(x_1^{e_1}, \dots, x_{k-1}^{e_{k-1}}, x_1^{d_1} x_2^{e_2 - d_2} \cdots x_{k-1}^{e_{k-1} - d_{k-1}}) = I' + (x_1^{d_1} x_2^{e_2 - d_2} \cdots x_{k-1}^{e_{k-1} - d_{k-1}}).
$$
Thus we may write $b_k$ as
$$
b_k = c_1 x_1^{e_1} + \cdots + c_{k-1} x_{k-1}^{e_{k-1}} + c x_1^{d_1} x_2^{e_2 - d_2} \cdots x_{k-1}^{e_{k-1} - d_{k-1}}.
$$
Plugging this back into equation (1), we may absorb the terms involving $c_1 x_1^{e_1}, \dots, c_{k-1} x_{k-1}^{e_{k-1}}$ into the other terms in the right-hand side of (1), and thereby assume without loss of generality that
$$b_k = c x_1^{d_1} x_2^{e_2 - d_2} \cdots x_{k-1}^{e_{k-1} - d_{k-1}}.$$
Then equation (1) says:
$$
a x_1^{d_1} x_2^{e_2 - d_2} \cdots x_k^{e_k - d_k} = b_1 x_1^{e_1} + \cdots + b_{k-1} x_{k-1}^{e_{k-1}} + c x_1^{d_1} x_2^{e_2 - d_2} \cdots x_{k-1}^{e_{k-1} - d_{k-1}} x_k^{e_k},
$$
or after moving the last term to the left,
$$
(a - c x_k^{d_k}) x_1^{d_1} x_2^{e_2 - d_2} \cdots x_k^{e_k - d_k} = b_1 x_1^{e_1} + \cdots + b_{k-1} x_{k-1}^{e_{k-1}}.
$$
We have now succeeded in removing the last term of equation (1)--at the cost of adjusting $b_1, \dots, b_{k-1}$, and adjusting $a$ by $-c x_k^{d_k}$. Since $r = ax_2^{e_2 - d_2} \cdots x_k^{e_k - d_k}$, this has the effect of adjusting $r$ by
$$-cx_2^{e_2 - d_2} \cdots x_{k-1}^{e_{k-1} - d_{k-1}} \cdot x_k^{e_k},$$
which lies in $(x_k^{e_k})$ and thus in $K$. So there is no harm in renaming $a - c x_k^{d_k}$ as $a$ and similarly for $r$; we must still show that $r$ lies in $K$.
Step 3: finishing up
At this point, the modified equation (1) says that
$$
a x_1^{d_1} x_2^{e_2 - d_2} \cdots x_k^{e_k - d_k} \in I' = (x_1^{e_1}, \dots, x_{k-1}^{e_{k-1}}).
$$
Since $x_k^{e_k-d_k}$ is a non-zero-divisor mod $I'$, we may cancel it from the product on the left-hand side:
$$
a x_1^{d_1} x_2^{e_2 - d_2} \cdots x_{k-1}^{e_{k-1} - d_{k-1}} \in I'.
$$
We can simplify this further using some degenerate cases of our inductive hypothesis: the element
$$
a x_2^{e_2 - d_2} \cdots x_{k-1}^{e_{k-1} - d_{k-1}}
$$
lies in the colon ideal
$$
I' : (x_1^{d_1}) = I' : (x_1^{d_1}, x_2^{e_2}, \dots, x_{k-1}^{e_{k-1}}) = I' : I_1',
$$
which by the inductive hypothesis equals
$$
I' + (x_1^{e_1 - d_1}) = (x_1^{e_1 - d_1}, x_2^{e_2}, \dots, x_{k-1}^{e_{k-1}}).
$$
Repeating this argument to eliminate powers of $x_2, \dots, x_{k-1}$ in turn ultimately implies that
$$
a \in (x_1^{e_1 - d_1}, x_2^{d_2}, \dots, x_{k-1}^{d_{k-1}}),
$$
and thus $r = ax_2^{e_2 - d_2} \cdots x_k^{e_k - d_k} \in K$, as claimed.
