Let me redo things with $f(x) = \frac{\delta^{-k}}{k!} \Delta_\delta^k g(x-k\delta)$, where $g(x) = (1-x)^k$ for $x\leq 1$ and $g(x)=0$ for $x>1$, and $\Delta h = h(y+\delta) - h(y)$. It is almost identical to the previous choice (with the multiplicative difference $\Delta_y^\cdot$) for $y = 1 + \delta$, $\delta$ small, and might be slightly more natural for the problem as stated. (Arguably, the version with $\Delta_y^\cdot$ is more natural for the problem that has $Mf(it)$ in place of $Mf(1+i t)$, not that that makes a big difference.) Some things become simpler and others more complicated.
I will give a variant at the end with $f(x) = \frac{\delta^{-k}}{k!} \Delta_\delta^k g(y-k\delta/2)$ (valid if we drop the condition that $f(x)=1$ for all $0\leq x\leq 1$); it gives better results.
It should be clear that $f(x)=1$ for $0\leq x\leq 1$ and $f(x)=0$ for $x\geq y + k \delta$. We can write $\Delta_\delta^k g(y-k\delta)$ as the convolution of $g_k$ with $ 1_{[0,\delta]}^{* k}= 1_{[0,\delta]}\ast \dotsc \ast 1_{[0,\delta]}$ ($k$ times), where $g_k(x) = g^{(k)}(x) = k!$ for $x\leq 1$ and $g_k(x) = 0$ for $x>1$. (Thanks to Noam Elkies for this suggestion.) It is then clear that $\Delta_\delta^k g(x-k\delta)$ is decreasing for $1< x< 1+k\delta$ (since $g_k$ is non-increasing). Hence, $f(x)$ is decreasing for $1<x\leq 1 + k \delta$, going from $f(1)=1$ to $f(1+k\delta)=0$. Since $1_{[0,\delta]}^{*k}$ is symmetric around $\delta k/2$ and of integral $\delta^k$, and the value at $x\geq 1$ of the convolution of $g_k$ with $1_{[0,\delta]}^{*k}$ can be written as $k!$ times the integral of $1_{[0,\delta]}^{*k}$ from $0$ to $x-1$, we see that $\frac{\delta^{-k}}{k!} \Delta_{\delta}^k (1+t) = 1- \frac{\delta^{-k}}{k!} \Delta_{\delta}^k (1+k\delta - t)$ for $0\leq t\leq k\delta$, and so $f(1 + t) = 1-f(1+k\delta - t)$ for $0\leq t\leq k\delta$. It folows that $$|f-1_{[0,1]}|_1 = \int_1^\infty f(x) dx = \frac{k \delta}{2}.$$
Taking the Mellin transform of $f(x)$ is easy: $$\begin{aligned}Mf(s) &= \frac{\delta^{-k}}{k!} \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} M(g(x - (k-j) \delta))(s) \\ &= \delta^{-k} \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} \frac{(1+ (k-j) \delta)^{s+k}}{s (s+1) \dotsc (s+k)}. \end{aligned}$$ The derivative of $\frac{(1+ (k-j) \delta)^{s+k}}{s (s+1) \dotsc (s+k)}$ equals $$\frac{(1+ (k-j) \delta)^{s+k}}{s (s+1) \dotsc (s+k)} \left(\log\left(1 + (k-j)\delta\right) - \frac{1}{s} - \frac{1}{s+1} - \dotsc - \frac{1}{s+k}\right)$$
Since $t\mapsto t^{k+1}$ is convex-up, $$\begin{aligned} \left|\sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} (1+ (k-j) \delta)^{s+k}\right|&\leq \sum_{j=0}^k \binom{k}{j} (1+(k-j) \delta)^{k+1}\leq 2^k \left(1 + \frac{k}{2} \delta\right)^{k+1} \end{aligned}$$ and similarly
In other words, $Mf(s) = \frac{C_{k}(s)}{s (s+1) \dotsc (s+k)}$, where $$|C_{k}(s)|\leq \left(\frac{2}{\delta}\right)^k \left(1 + \frac{k}{2} \delta\right)^{k+1} \sim \left(\frac{2}{\delta}\right)^k.$$ Hence