We prove by induction on $k$ that
$$\begin{equation*} u_{k}=\sum_{\substack{0<i_{1}<\ldots<i_{t}<k \\ i_{j+1}-i_{j} \geqslant 2}} a_{i_{1}} \ldots a_{i_{t}} \tag{1}\end{equation*}$$
Note that we have one trivial summand equal to 1 (which corresponds to $t=0$ and the empty sequence, whose product is 1).
For $k=0,1$ the sum on the right-hand side only contains the empty product, so (1) holds due to $u_{0}=u_{1}=1$. For $k \geqslant 1$, assuming the result is true for $0,1, \ldots, k$, we have
$$\begin{aligned} u_{k+1} & =\sum_{\substack{0<i_{1}<\ldots<i_{t}<k, i_{j+1}-i_{j} \geqslant 2}} a_{i_{1}} \ldots a_{i_{t}}+\sum_{\substack{0<i_{1}<\ldots<i_{t}<k-1, i_{j+1}-i_{j} \geqslant 2}} a_{i_{1}} \ldots a_{i_{t}} \cdot a_{k} \\ & =\sum_{\substack{0<i_{1}<\ldots<i_{t}<k+1, i_{j+1}-i_{j} \geqslant 2, k \notin\left\{i_{1}, \ldots, i_{t}\right\}}} a_{i_{1}} \ldots a_{i_{t}}+\sum_{\substack{0<i_{1}<\ldots<i_{t}<k+1, i_{j+1}-i_{j} \geqslant 2, k \in\left\{i_{1}, \ldots, i_{t}\right\}}} a_{i_{1}} \ldots a_{i_{t}} \\ & =\sum_{\substack{0<i_{1}<\ldots<i_{t}<k+1, i_{j+1}-i_{j} \geqslant 2}} a_{i_{1}} \ldots a_{i_{t}},\end{aligned}$$
as required.
Applying (1) to the sequence $b_{1}, \ldots, b_{n}$ given by $b_{k}=a_{n-k}$ for $1 \leqslant k \leqslant n$, we get
$$\begin{equation*} v_{k}=\sum_{\substack{0<i_{1}<\ldots<i_{t}<k \\ i_{j+1}-i_{j} \geqslant 2}} b_{i_{1}} \ldots b_{i_{t}}=\sum_{\substack{n>i_{1}>\ldots>i_{t}>n-k \\ i_{j}-i_{j+1} \geqslant 2}} a_{i_{1}} \ldots a_{i_{t}} \tag{2}\end{equation*}$$
For $k=n$ the expressions (1) and (2) coincide, so indeed $u_{n}=v_{n}$.

Solution 2:
Define recursively a sequence of multivariate polynomials by
$$P_{0}=P_{1}=1, \quad P_{k+1}\left(x_{1}, \ldots, x_{k}\right)=P_{k}\left(x_{1}, \ldots, x_{k-1}\right)+x_{k} P_{k-1}\left(x_{1}, \ldots, x_{k-2}\right),$$
so $P_{n}$ is a polynomial in $n-1$ variables for each $n \geqslant 1$. Two easy inductive arguments show that
$$u_{n}=P_{n}\left(a_{1}, \ldots, a_{n-1}\right), \quad v_{n}=P_{n}\left(a_{n-1}, \ldots, a_{1}\right)$$
so we need to prove $P_{n}\left(x_{1}, \ldots, x_{n-1}\right)=P_{n}\left(x_{n-1}, \ldots, x_{1}\right)$ for every positive integer $n$. The cases $n=1,2$ are trivial, and the cases $n=3,4$ follow from $P_{3}(x, y)=1+x+y$ and $P_{4}(x, y, z)= 1+x+y+z+x z$.
Now we proceed by induction, assuming that $n \geqslant 5$ and the claim hold for all smaller cases. Using $F(a, b)$ as an abbreviation for $P_{|a-b|+1}\left(x_{a}, \ldots, x_{b}\right)$ (where the indices $a, \ldots, b$ can be either in increasing or decreasing order),
$$\begin{aligned} F(n, 1) & =F(n, 2)+x_{1} F(n, 3)=F(2, n)+x_{1} F(3, n) \\ & =\left(F(2, n-1)+x_{n} F(2, n-2)\right)+x_{1}\left(F(3, n-1)+x_{n} F(3, n-2)\right) \\ & =\left(F(n-1,2)+x_{1} F(n-1,3)\right)+x_{n}\left(F(n-2,2)+x_{1} F(n-2,3)\right) \\ & =F(n-1,1)+x_{n} F(n-2,1)=F(1, n-1)+x_{n} F(1, n-2) \\ & =F(1, n) \end{aligned}$$
as we wished to show.

Solution 3:
Using matrix notation, we can rewrite the recurrence relation as
$$\binom{u_{k+1}}{u_{k+1}-u_{k}}=\binom{u_{k}+a_{k} u_{k-1}}{a_{k} u_{k-1}}=\left(\begin{array}{cc} 1+a_{k} & -a_{k} \\ a_{k} & -a_{k} \end{array}\right)\binom{u_{k}}{u_{k}-u_{k-1}}$$
for $1 \leqslant k \leqslant n-1$, and similarly
$$\left(v_{k+1} ; v_{k}-v_{k+1}\right)=\left(v_{k}+a_{n-k} v_{k-1} ;-a_{n-k} v_{k-1}\right)=\left(v_{k} ; v_{k-1}-v_{k}\right)\left(\begin{array}{cl} 1+a_{n-k} & -a_{n-k} \\ a_{n-k} & -a_{n-k} \end{array}\right)$$
for $1 \leqslant k \leqslant n-1$. Hence, introducing the $2 \times 2$ matrices $A_{k}=\left(\begin{array}{cl}1+a_{k} & -a_{k} \\ a_{k} & -a_{k}\end{array}\right)$ we have
$$\binom{u_{k+1}}{u_{k+1}-u_{k}}=A_{k}\binom{u_{k}}{u_{k}-u_{k-1}} \quad \text{ and } \quad\left(v_{k+1} ; v_{k}-v_{k+1}\right)=\left(v_{k} ; v_{k-1}-v_{k}\right) A_{n-k} .$$
for $1 \leqslant k \leqslant n-1$. Since $\binom{u_{1}}{u_{1}-u_{0}}=\binom{1}{0}$ and $\left(v_{1} ; v_{0}-v_{1}\right)=(1 ; 0)$, we get
$$\binom{u_{n}}{u_{n}-u_{n-1}}=A_{n-1} A_{n-2} \cdots A_{1} \cdot\binom{1}{0} \quad \text{ and } \quad\left(v_{n} ; v_{n-1}-v_{n}\right)=(1 ; 0) \cdot A_{n-1} A_{n-2} \cdots A_{1} .$$
It follows that
$$\left(u_{n}\right)=(1 ; 0)\binom{u_{n}}{u_{n}-u_{n-1}}=(1 ; 0) \cdot A_{n-1} A_{n-2} \cdots A_{1} \cdot\binom{1}{0}=\left(v_{n} ; v_{n-1}-v_{n}\right)\binom{1}{0}=\left(v_{n}\right) .$$