Conjecture 1. Let $(x_n)_{n=1}^\infty$ be a (Schauder) basis for a Banach space $X$. Set $y_1=x_1$ and $y_n=x_n-x_{n-1}$ for $n\geq 2$. Then $(y_n)_{n=1}^\infty$ is a basis for $X$.
It is clear that $(y_n)_{n=1}^\infty$ spans $X$, so we need only show that $(y_n)_{n=1}^\infty$ is a basic sequence.
Note that if $N\geq 2$ then $\sum_{n=1}^Na_ny_n=\sum_{n=1}^{N-1}(a_n-a_{n+1})x_n+a_Nx_N$. Hence, for $2\leq M<N$ and scalars $(a_n)_{n=1}^N$ we have \begin{equation*}\|\sum_{n=1}^Ma_ny_n\|\leq K\|\sum_{n=1}^Na_ny_n\|+\|a_Mx_M\|,\end{equation*} where $K$ is the basis constant for $(x_n)_{n=1}^\infty$. However, this is not enough to prove that $(y_n)_{n=1}^\infty$ is a basis.
Conjecture 2. If furthermore $(x_n)_{n=1}^\infty$ is spreading then so is $(y_n)_{n=1}^\infty$.
Here, a basic sequence is said to be spreading if there is $C\in[1,\infty)$ such that it is $C$-equivalent to all its subsequences, i.e. if $(n_i)_{i=1}^\infty$ is a strictly increasing sequence of positive integers then for every $N\in\mathbb{N}$ and every finite sequence of scalars $(a_n)_{n=1}^N$, \begin{equation}C^{-1}\|\sum_{i=1}^Na_ix_{n_i}\|\leq\|\sum_{n=1}^Na_nx_n\|\leq C\|\sum_{i=1}^Na_ix_{n_i}\|\end{equation}
No. Suppose that $(y_n)$ is in fact a basis.
If $x=\sum_n c_nx_n$ let $$S_nx=\sum_{j=1}^n c_jx_j.$$
Now it's possible to write $$S_nx=\sum_{j=1}^na_jy_j.$$Since $(y_n)$ is a basis there's only one way to do this, and it follows that $$a_1=c_1+\dots+c_n.$$So there must exist $C$ so that $$|c_1+\dots +c_n|\le C||S_nx||.$$This certainly seems implausible; a simple example where it's clearly false is given by taking $(x_n)$ to be an orthonormal basis for a Hilbert space.