You are correct in saying that Tanaka's SDE
$$
\begin{align}
dX&={\rm sgn}(X)\,dB,\\
X_0&=0
\end{align}
$$
does not admit strong solutions, so that $X$ cannot be constructed as a function of $B$. As mentioned by TheBridge, the Brownian motion $B$ can be written as $\int{\rm sgn}(X)\,dX$, which is a function of $\vert X\vert$ (it is the martingale part of $\vert X\vert$) and, consequently, ${\rm sgn}(X_t)$ is independent of $B$.
However, the Euler-Maruyama method still works here. You do not obtain approximate solutions converging to a function of $B$, but you do get convergence in distribution. It is not too hard to show how this happens. Suppose that we are constructing solutions over an interval $[0,T]$. Then, for a positive integer $N$ you would construct an approximation $X^{(N)}$, say, by setting
$$
\begin{align}
&X^{(N)}_0=0,\\
&X^{(N)}_\frac{k+1}N=X^{(N)}_{\frac kN}+{\rm sgn}\left(X^{(N)}_{\frac kN}\right)\,\left(B_\frac{k+1}N-B_\frac kN\right)
\end{align}
$$
for $k=0,1,\ldots,N-1$. This defines the discrete approximation at times $k/N$. As the term ${\rm sgn}(X^{(N)}_\frac kN)$ is independent of $B_\frac{k+1}N-B_\frac kN$, the process $X$ has normally distributed independent increments. Actually, I think it is a bit easier to extend the definition of $X^{(N)}$ to all intermediate times $\frac kN\le t\le\frac{k+1}N$ by
$$
X^{(N)}_t=X^{(N)}_{\frac kN}+{\rm sgn}\left(X^{(N)}_{\frac kN}\right)\,\left(B_t-B_\frac kN\right),
$$
in which case $X^{(N)}$ is a Brownian motion.
This can be expressed conveniently as a stochastic integral
$$
X^{(N)}_t=\int_0^t{\rm sgn}(X_{\lfloor sN\rfloor/N})\,dB_s
$$
where $\lfloor\cdot\rfloor$ is the floor function. This does not converge, in probability, to a limit as $N\to\infty$. However, we do have
$$
B_t=\int_0^t{\rm sgn}(X^{(N)}_{\lfloor sN\rfloor/N})\,dX^{(N)}_s.
$$
So, if $\tilde X$ is any standard Brownian motion (defined on any probability space), we have equality of distributions
$$
\left(B,X^{(N)}\right)\sim\left(\int{\rm sgn}(\tilde X_{\lfloor sN\rfloor/N})\,d\tilde X_s,\tilde X\right).
$$
We do have convergence ${\rm sgn}(\tilde X_{\lfloor sN\rfloor/N})\to{\rm sgn}(\tilde X_s)$ outside of the (zero measure) set of times where $\tilde X_s=0$. Bounded convergence of stochastic integrals means that we get convergence in distribution
$$
\left(B,X^{(N)}\right)\xrightarrow{\rm d}\left(\tilde B,\tilde X\right)
$$
where $\tilde B=\int{\rm sgn}(\tilde X)\,d\tilde X$. Finally, $\tilde X$ does indeed satisfy Tanaka's SDE with respect to the driving Brownian motion $\tilde B$, but is not a function of $\tilde B$.
So, on an individual sample path for $B$, we do not have convergence of the paths of $X^{(N)}$ to a function of $B$, but do have convergence in distribution to some process which actually has additional randomness not contained in the paths of $B$. What is happening is that, when $X^{(N)}$ becomes small, ${\rm sgn}(X^{(N)})$ takes the values $1$ and $-1$ with roughly equal probability. Only a small change in $B$ (or small change in the time when $B$ is sampled) will reverse the sign of ${\rm sgn}(X^{(N)}_s)$. So, as $N$ becomes large, ${\rm sgn}(X^{(N)}_s)$ becomes independent of $B$ in the limit. You do get pathwise convergence of $\vert X^{(N)}\vert$. In fact
$$
\vert X^{(N)}_t\vert\to B_t-\min_{s\le t}B_s
$$
but ${\rm sgn}(X^{(N)})$ moves around at random, and only converges in distribution.
Start with a Brownian motion $X$ carried by some probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and define $W_t = \int_0^t \mathrm{sgn}(X_s)dX_s$. By Lévy's characterisation theorem, $W$ is a Brownian motion. Observe then that by associativity of the stochastic integral: $$\mathrm{sgn}(X_s) dW_s = \mathrm{sgn}(X_s)^2 dX_s = dX_s$$
showing that $(X,W)$ is a weak solution of your proposed SDE.
Best Answer
So my original answer did not take into account the fact that there could exist a transformation of the entire path $(|X_{s}|)_{0\leq s\leq t}$. Instead, I think I have found a way to prove that for a Brownian motion $X$, we cannot have the inclusion $\mathcal{F}_{t}^{X}\subseteq\mathcal{F}_{t}^{|X|}$. Since the solution to Tanaka's SDE is a Brownian motion, this will suffice.
Define the $\sigma$-algebra $$ \mathcal{D}_{t} = \lbrace F\in \mathcal{F}_{t}^{|X|}\mid \mathbb{E}\left[ X_{t}\mathbb{1}_{F} \right] = 0 \rbrace $$ That this is in fact a $\sigma$-algebra follows by linearity of the integral and that $X_{t}\in L^{1}$ with expectation 0.
Let $B\in\mathcal{B}(\mathbb{R})$ and $s\leq t$. Then $\lbrace |X_{s}|\in B\rbrace \in \mathcal{D}_{t}$ since by the reflection principle $$ \mathbb{E}\left[ X_{t}\mathbb{1}_{\lbrace |X_{s}|\in B\rbrace } \right] = 0 $$ This means that $\mathcal{G}_{t}=\mathcal{F}_{t}^{|X|}$, at least if we assume the filtration $\mathcal{F}_{t}^{|X|}$ to be complete.
But since $$ \mathbb{E}\left[ X_{t}\mathbb{1}_{\lbrace X_{t}>0\rbrace} \right] > 0 $$ we have $\lbrace X_{t}>0\rbrace\notin\mathcal{F}_{t}^{|X|}$, but naturally we have $\lbrace X_{t}>0 \rbrace\in\mathcal{F}_{t}^{X}$ and thus we must have $$ \mathcal{F}_{t}^{X}\not\subseteq\mathcal{F}_{t}^{|X|} $$ as desired.
This lines up with the intuition that knowing something about $|X_{t}|$ does not tell us about the sign of $X_{t}$.