This is an answer to the second version: a function $f$ such that $f(⋅−k)$ form an orthonormal basis with the same span as $(g(⋅−λ_k))$ exists only in very special cases.
Assume it exists, then on the Fourier transform side span $(g-\lambda_k)$ is the space of functions of the form $\phi\cdot \hat f$ where $\phi$ is $2\pi$-periodic, so in particular we have $e^{i\lambda_k t}\hat g(t)=\phi_k(t)\hat f(t)$. Let $S$ be the support of $\hat g$. Clearly $\hat f$ is non zero on $S$,  so on $S$ we have $e^{i\lambda_k t}=\phi_k(t)[\hat f(t)/\hat g(t)]$. Writing this for integers $k\neq s$ and dividing we see that $e^{i(\lambda_k-\lambda_s)t}$ is $2\pi$ periodic on $S$. If the measure of $S$ is $>2\pi$ there are two sets of positive measure $A$ and $B$ such that $A=2\pi \mu +B$ for some $\mu\in Z$. Thus for $t\in B$ we have $e^{i(\lambda_k-\lambda_s)(t+2\pi \mu)}=e^{i(\lambda_k-\lambda_s)t}$ which gives $e^{i(\lambda_k-\lambda_s)2\pi \mu}=1$ so $\lambda_k-\lambda_s$ is an integer for all $k,s\in Z$ which implies $\lambda_k=k+\delta$.
If measure of $S$ is  $\leq2\pi$ then such $f$ can exist only if $\sum_{s\in Z}I(t+2\pi s t)=1$ for $I$ an indicator function of $S$.

For the first version here is a counterexample. Let $g$ be the indicator function of the interval $[-1/2,1/2]$. Then $(g(x-k))$ is an orthonormal basis so a Riesz basis. Let $\lambda_k=k+\delta_k$ with $1/3>\delta_k>\delta_{k+1}>0$. Denote $g_j(x)=:g(x-\lambda_j)$. It is a Riesz basis. We have $<g_j,g_s>$ equals $1$ if $j=s$ and equals $0$ if $|j-s|>1$. Otherwise it is small, so the Gram matrix is diagonally dominated, so invertible. 
If $f$ belongs to span of $(g_j)$ then $f=\sum_{j\in Z} x_j g_j$ with $(x_j)$ square summable. We see that $x_j=f(j)$. Take $f\neq 0$ and for simplicity assume that $x_0=f(0)=\alpha\neq 0$. The function $f$ equals $0$ on the interval $(-1/2-\delta_{-1}; -1/2 -\delta_0)$, equals $\alpha$ on the interval $(-1/2-\delta_0; 1/2+\delta_1)$ and $\alpha+x_1$ on the interval $(1/2+\delta_1;1/2 +\delta_0)$. Note that if we consider $f(x-\lambda_0)= f(x-\delta_0)$ we get a step functions but with different intervals. Thus there is no non-zero function $f$ in span of $(g_j)$ such that the whole sequence $(f(x-\lambda_j))$ is in this span.