*I originally saw the idea that inspired me to this answer here*.

From the statement it follows that

$$
g_k(t_1, t_2, \dots, t_k) - g_k(t_1 - 1, t_2, \dots, t_k) = g_{k-1}(t_2+t_1, t_3+t_1, \dots, t_k+t_1).
$$

Generally, one can express shift in the argument of analytic function with the operator exponent

$$
f(x+a) = e^{a\frac{\partial}{\partial x}}f(x),
$$

Let's denote $D_k = \frac{\partial}{\partial t_k}$, then the recurrence above rewrites as

$$
(1-e^{-D_1}) g_k = e^{t_1 (D_2+D_3+\dots + D_k)}g_{k-1},
$$

We can't invert $(1-e^{-D_1})$ operator directly because it doesn't distinguish between $0$ and $1$.

However we know that $g_k(0, t_2, \dots, t_k) = g_{k-1}(t_2, \dots, t_k)$, hence we can just solve the equation above for higher degrees of $t_1$, meaning that it's sufficient to find $D_1 g_k$:

$$
D_1 g_k = \frac{D_1}{1-e^{-D_1}}e^{t_1(D_2+\dots+D_k)} g_{k-1}
$$

Now let $I_k$ be the integrating operator over $t_k$ such that $[t_k^0]I_k f(t_k, \dots)=0$, we can express $g_k$ as

$$
g_k(t_1, t_2, \dots, t_k) = \left[1 + I_1 \cdot \frac{D_1}{1-e^{-D_1}} \cdot e^{t_1(D_2+\dots+D_k)} \right]g_{k-1}(t_2, \dots, t_k).
$$

Substituting it for $g_{k-1}$ and up to the very end, we get

$$
g_k(t_1, t_2, \dots, t_k) = \prod\limits_{j=1}^k \left(1+I_j \cdot \frac{D_j}{1-e^{-D_j}} \cdot e^{t_j(D_{j+1}+\dots+D_k)} \right) \cdot 1,
$$

assuming that the product unravels left-to-right as $j$ grows and then is applied to the constant $1$.

### Example for $k=1$

The expression in this case is simply

$$
g_1(t_1) = \left(1+I_1 \frac{D_1}{1-e^{-D_1}}\right) \cdot 1
$$

The exponent operator rewrites as

$$
\frac{D_1}{1-e^{-D_1}} = \frac{1}{\sum\limits_{k=1}^\infty \frac{(-D_1)^{k-1}}{k!}} = \frac{1}{1-\frac{D_1}{2}+O(D_1^2)} = 1+\frac{D_1}{2}+O(D_1^2).
$$

When it is applied to $1$, it yields $1$, then $I_1 \cdot 1 = t_1$, as the integration makes sure that it is $0$ for $t_1=0$.

Then, adding $1$ the result is, indeed, $g_1 = 1+t_1$.