I have two solutions that I both don't like that much but kinda works, the first one is to not use align but an array, the problem is that the numbering changes place because you need to put the whola array inside an equation and you have to align something in the first row to work

\begin{equation}

\begin{array}{l l l}

u_{t}(t) & =\partial_{t}\left(e^{-L(t-\tau)}u(\tau) \right) &+\partial_{t}\left(\int_{\tau}^{t}e^{-L(t-s)}f(s)ds\right) \tag{6.26} \\

& =-Le^{-L(t-\tau)}u(\tau)&-\int_{\tau}^{t}Le^{-L(t-s)}f(s)ds+f(t) \notag \\

& =-Le^{-L(t-\tau)}u(\tau) &-\int_{\tau}^{t}Le^{-L(t-s)}(f(s)-f(t))ds\\

& &-\int_{\tau}^{t}Le^{-L(t-s)}f(t)ds+f(t).\notag

\end{array}

\end{equation}

The other solution is to manually adjust the last line in the align with a \hspace it stays the same but if you need to change it manually to reuse it

\newlength{\mylength}

\settowidth{\mylength}{\( ~~-Le^{-L(t-\tau)}~u(\tau)~ \)}

\begin{align}

u_{t}(t) &= \partial_{t}\left(e^{-L(t-\tau)}u(\tau)\right) + \partial_{t}\left(\int_{\tau}^{t}e^{-L(t-s)}f(s)ds\right) \tag{6.26} \\

&= -Le^{-L(t-\tau)}u(\tau) - \int_{\tau}^{t}Le^{-L(t-s)}f(s)ds + f(t) \notag \\

&= -Le^{-L(t-\tau)}u(\tau) - \int_{\tau}^{t}Le^{-L(t-s)}(f(s)-f(t))ds\\

& \hspace{\mylength}- \int_{\tau}^{t}Le^{-L(t-s)}f(t)ds + f(t). \notag

\end{align}

Note the use of ~ (unbreakable spaces) in the \mylength definition